RD Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 12 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among class 12 Science students for Math Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 2 2018 Book of class 12 Science Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 2 2018 Solutions. All RD Sharma XII Vol 2 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 31.102:

Question 1:

A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5.

Answer:

To be divisible by 5 ones place sholud be 5There are 3 places remaining which can be filled in 3 !=6 waysSo, 6 numbers can be formed out of 1, 2, 3 and 5, which are divisible by 5.Total 4-digit numbers=4!=24P4-digit number divisible by 5=624=14

Page No 31.102:

Question 2:

When three dice are thrown, write the probability of getting 4 or 5 on each of the dice simultaneously.

Answer:

P4 or 5 on a die=26=13Pgetting 4 or 5 on each of the dice simultaneously=13×13×13=127

Page No 31.102:

Question 3:

Three digit numbers are formed with the digits 0, 2, 4, 6 and 8. Write the probability of forming a three digit number with the same digits.

Answer:

Total 3-digit numbers that can be made out of 0, 2, 4, 6 and 8=4×5×5                     hundreds place cannot be filled with 0=100But 222, 444, 666 and 888 are four numbers, which have the same digits at all places.P3-digit number having same digits at all places=4100=125



Page No 31.103:

Question 4:

A ordinary cube has four plane faces, one face marked 2 and another face marked 3, find the probability of getting a total of 7 in 5 throws.

Answer:

A cube has total 6 faces.Total possible outcomes in 5 throws=6×6×6×6×6=65The only way of getting 7 is by getting two 2s and one 3.Total possible ways=P532!=5×4×3×2×12×1×2×1=30Now,Pgetting 7 in 5 throws=3065=564

Page No 31.103:

Question 5:

Three numbers are chosen from 1 to 20. Find the probability that they are consecutive.

Answer:

Total possible outcomes=20C3Consecutive numbers chosen=(1, 2, 3), (2, 3, 4) ... (18, 19, 20)So, there are 18 favouable cases.PA=18C203

Page No 31.103:

Question 6:

6 boys and 6 girls sit in a row at random. Find the probability that all the girls sit together.

Answer:

Here, 6 boys and 6 girls can be arranged in a line in 12! ways.Total possible outcomes=12!Consider 6 girls as a single element X.Now, 6 boys and X can be arranged in a line in 7! ways and girls can be arranged in 6! ways among them.Pall girls are together=7!×6!12!=7×6×5×4×3×2×1×6×5×4×3×2×112×11×10×9×8×7×6×5×4×3×2×1=111×12=1132

Page No 31.103:

Question 7:

If A and B are two independent events such that P (A) = 0.3 and P (AB) = 0.8. Find P (B).

Answer:

A and B are two independent events. PAB¯=PA+PB¯-PAB¯0.8=0.3+1-PB-PA PB¯0.5=1-PB-0.31-PB0.5=1-PB-0.3+O.3PB0.5=0.7-PB1-0.30.7PB=0.2PB=0.20.7=27

Page No 31.103:

Question 8:

An unbiased die with face marked 1, 2, 3, 4, 5, 6 is rolled four times. Out of 4 face values obtained, find the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5.

Answer:

Pface value is not more than 5 and not less than 2=46=23Pface value is not more than 5 and not less than 2 in 4 throws=23×23×23×23=1681

Page No 31.103:

Question 9:

If A and B are two events write the expression for the probability of occurrence of exactly one of two events.

Answer:

Pexactly one of 2 events=PAB-PAB=PA+PB-PAB-PAB=PA+PB-2PAB

Page No 31.103:

Question 10:

Write the probability that a number selected at random from the set of first 100 natural numbers is a cube.

Answer:

Number of cubes in first 100 natural numbers = 1,8,27,64So, there are 4 cubes in first 100 natural numbers.Pgetting a cube from a set of first 100 natural numbers=4100=125

Page No 31.103:

Question 11:

In a competition A, B and C are participating. The probability that A wins is twice that of B, the probability that B wins is twice that of C. Find the probability that A losses.

Answer:

PA wins=PAPB wins=PB=PA2PC wins=PC=PB2=PA4Now,PA+PB+PC=1PA+PA2+PA4=1PA4+2+14=1PA=47PA loses=PA¯=1-47=37

Page No 31.103:

Question 12:

If A, B, C are mutually exclusive and exhaustive events associated to a random experiment, then write the value of P (A) + P (B) + P (C).

Answer:

A, B and C are mutually exclusive and exhaustive events. PA+PB+PC=1

Page No 31.103:

Question 13:

If two events A and B are such that P A = 0.3, P (B) = 0.4 and P (AB) = 0.5, find P B/AB.

Answer:

Disclaimer: the question seems to be incorrect.

Page No 31.103:

Question 14:

If A and B are two independent events, then write P (AB) in terms of P (A) and P (B).

Answer:

A and B are two independent events. PAB¯=PAPB¯=PA1-PB=PA-PAPB

Page No 31.103:

Question 15:

If P (A) = 0.3, P (B) = 0.6, P (B/A) = 0.5, find P (AB).

Answer:

PBA=PABPA0.5=PAB0.3PAB=0.5×0.3PAB=0.15Now,PAB=PA+PB-PAB             =0.3+0.6-0.15            =0.9-0.15            =0.75

Page No 31.103:

Question 16:

If A, B and C are independent events such that P(A) = P(B) = P(C) = p, then find the probability of occurrence of at least two of A, B and C.

Answer:

PAt least two of A, B and C occur=PExactly two of A, B and C occurs+PAll three occurs=PAB-PABC+PBC-PABC+PAC-PABC+PABC=PAB+PBC+PAC-3PABC+PABC=PAB+PBC+PAC-2PABC=PA×PB+PB×PC+PA×PC-2PA×PB×PC          As, A, B and C are independent events=p×p+p×p+p×p-2p×p×p=p2+p2+p2-2p3=3p2-2p3

So, the probability of occurrence of at least two of AB and C 3p2-2p3.

Disclaimer: The answer given in the book is incorrect. the same has been corrected above.

Page No 31.103:

Question 17:

If A and B are independent events, then write expression for P(exactly one of AB occurs).

Answer:

As, A and B are independent events.So, PAB=PA×PB            .....iNow,Pexactly one of A, B occurs=Ponly A+Ponly B=PA-PAB+PB-PAB=PA-PA×PB+PB-PA×PB                  Using i=PA×1-PB+PB×1-PA=PA×PB¯+PB×PA¯

Page No 31.103:

Question 18:

If A and B are independent events such that P(A) = p, P(B) = 2p and P(Exactly one of A and B occurs) = 59, then find the value of p.

Answer:

As, A and B are independent events.So, PAB=PA×PB              .....iNow,PExactly one of A and B occurs=59POnly A+POnly B=59PA-PAB+PB-PAB=59PA-PA×PB+PB-PA×PB=59              Using iPA×1-PB+PB×1-PA=59p1-2p+2p1-p=59p-2p2+2p-2p2=593p-4p2=5927p-36p2=536p2-27p+5=0So, using quadratic formula, we getp=--27±-272-4×36×52×36=27±729-72072=27±972=27±372p=27+372 or p=27-372p=3072 or p=2472 p=512 or p=13

Page No 31.103:

Question 1:

If one ball is drawn at random from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, then the probability that 2 white and 1 black balls will be drawn is
(a) 1332

(b) 14

(c) 132

(d) 316

Answer:

a 13 32Here, the three boxes contain 3 white and1black 3W, 1B, 2 white and 2 black 2W, 2B and 1 white and 3 blackballs1W, 3B, respectively.P2W, 1B=34×24×34+34×24×14+14×24×14=1864+664+264=2664=1332

Page No 31.103:

Question 2:

A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is
(a) 4485×49

(b) 1185×49

(c) 13×2417×25×49

(d) none of these

Answer:

a 4485×49Total cards=52There are four suits of cards in a pack, i.e. diamond, heart, spade and club.Pall 4 cards are of same suit=Pall 4 cards are of diamond+Pall 4 cards are of heart+Pall 4 cards are of spade+Pall 4 cards are of club=4×1352×1251×1150×1049=4×1185×49=4485×49

Page No 31.103:

Question 3:

A and B are two events such that P (A) = 0.25 and P (B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is
(a) 0.39
(b) 0.25
(c) 0.11
(d) none of these

Answer:

a  0.39PAB=PA+PB-PAB=0.25+0.5-0.14=0.61Pboth A and B not happening=PAB'=1-PAB=1-0.61=0.39



Page No 31.104:

Question 5:

India play two matches each with West Indies and Australia. In any match the probabilities of India getting 0,1 and 2 points are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is
(a) 0.0875
(b) 1/16
(c) 0.1125
(d) none of these

Answer:

a 0.0875Here, there are total 5 ways by which India can get at least 7 points. 1 2 points+ 2 points + 2 points + 2 points=0.5×0.5×0.5×0.52 1 point+ 2 points + 2 points + 2 points = 0.05×0.5×0.5×0.53 2 points+ 1 point + 2 points + 2 points = 0.5×0.05×0.5×0.54 2 points+ 2 points + 1 point + 2 points= 0.5×0.5×0.05×0.55 2 points+ 2 points + 2 points + 1 point = 0.5×0.5×0.5×0.05Patleast 7 points=0.5×0.5×0.5×0.5 +40.05×0.5×0.5×0.5=0.0625 +40.00625=0.0625+0.025=0.0875

Page No 31.104:

Question 6:

Three faces of an ordinary dice are yellow, two faces are red and one face is blue. The dice is rolled 3 times. The probability that yellow red and blue face appear in the first second and third throws respectively, is
(a) 136

(b) 16

(c) 130

(d) none of these

Answer:

a 136Pyellow face=36=12Pred face=26=13Pone face=16Pyellow face, red face and blue face appear in the required order=12×13×16=136

Page No 31.104:

Question 7:

The probability that a leap year will have 53 Fridays or 53 Saturdays is
(a) 27

(b) 37

(c) 47

(d) 17

Answer:

b 37

A leap year has 366 days 

For a non-leap year:
52 weeks + 1 day

For a  leap year:
52 weeks + 2 days

Sample space=Monday, Tuesday, Tuesday, Wednesday,Wednesday, Thursday,Thursday, Friday, Friday, Saturday, Saturday, Sunday, Sunday, MondayFavourable cases=3P53 Fridays or 53 Saturdays=37

Page No 31.104:

Question 8:

A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is
(a) 14

(b) 1124

(c) 1524

(d) 2324

Answer:

d 2324

4 letters can be placed in 4 envelopes in 4! ways = 24 ways
Now, there is only one method, by which all the letters are placed in the right envelope.

P(all letters are placed in the right envelopes) = 124 

P(all letters are not placed in the right envelopes) = 1 - P(all letters are placed in the right envelopes)

                                                                               =1-124=2324

Page No 31.104:

Question 9:

A speaks truth in 75% cases and B speaks truth in 80% cases. Probability that they contradict each other in a statement, is
(a) 720

(b) 1320

(c) 35

(d) 25

Answer:

a 720PA speaks truth=0.75PA lies= 1-0.75=0.25PB speaks truth=0.8PB lies= 1-0.8=0.2Pcontradicting each other in a statement=P(A speaks truth and B lies)+PB speaks truth and A lies=0.75×0.2+0.8×0.25=0.15+0.2=0.35=35100=720

Page No 31.104:

Question 10:

Three integers are chosen at random from the first 20 integers. The probability that their product is even is
(a) 219

(b) 329

(c) 1719

(d) 419

Answer:

c 1719Pproduct is even=1-Pproduct is oddFor the product to be odd, the two digits must be odd.Now, 10 numbers are odd in the first 20 integers. Pproduct is even=1-1020×919×818=1-938×818=1-219=1719

Page No 31.104:

Question 11:

Out of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd, is
(a) 1429

(b) 1629

(c) 1529

(d) 1029

Answer:

(c) 1529

For sum of two integers to be odd, one integer should be even and the other should be odd.
In 30 consecutive integers, 15 are even and 15 are odd.

P(sum is odd) = P(first integer is odd and second is even) + P(first integer is even and second integer is odd)

                       =1530×1529+1530×1529=45030×29=1529

Page No 31.104:

Question 12:

A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected randomwise, the probability that it is black or red ball is
(a) 13

(b) 14

(c) 512

(d) 23

Answer:

d 23We know that the bag contains 5B (black), 4W (white) and 3R (red)balls.Now,PB=512PR=312PB or R=PB+PR                 =512+312                 =812=23

Page No 31.104:

Question 13:

Two dice are thrown simultaneously. The probability of getting a pair of aces is
(a) 136

(b) 13

(c) 16

(d) none of these

Answer:

a 136Ppair of aces=Pace in first throw×Pace in second throw=16×16=136

Page No 31.104:

Question 14:

An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is
(a) 584

(b) 39

(c) 37

(d) 717

Answer:

a 584Given:Red balls=2Blue balls=3Black balls=4Pall three balls are of same colour=P(all three are blue)+Pall three are black=39×28×17+49×38×27=184+484=584

Page No 31.104:

Question 15:

A coin is tossed three times. If events A and B are defined as A = Two heads come, B = Last should be head. Then, A and B are
(a) independent
(b) dependent
(c) both
(d) mutually exclusive

Answer:

b dependentS=HHH, HHT, HTH, HTT, THH, THT, TTH, TTTPA=P2 heads=38PB=Plast one is head=48PAB=28=14PA PBThus, A and B are dependent.

Page No 31.104:

Question 4:

The probabilities of a student getting I, II and III division in an examination are 110,35and14 respectively. The probability that the student fails in the examination is
(a) 197200

(b) 27100

(c) 83100

(d) none of these

Answer:

b 27100Pstudent gets first division=110Pstudent gets second division=35Pstudent gets third division=14Pstudents fails =Pstudent does not get first division×Pstudent does not get second division×Pstudent does not get third division                      =1-1101-351-14                      =910×25×34                      =54200                      =27100



Page No 31.105:

Question 16:

Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floors is
(a) P5775

(b) 75P57

(c) 6P56

(d) P5555

Answer:

a P5775Total possible ways of leaving the lift =7×7×7×7×7=755 people can leave different floors in P57 ways.P5 people leaving the lift at different floors=P5775

Page No 31.105:

Question 17:

A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is
(a) 6464

(b) 4964

(c) 4064

(d) 2464

Answer:

a 6464Pgood item=1016Pdefected item =616Peither good or defected item=Pgood item+Pdefected item=1016+616=1616= 1=6464

Page No 31.105:

Question 18:

A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, the probability that it is rusted or is a nail is

(a) 316

(b) 516

(c) 1116

(d) 1416

Answer:

(c) 1116


Rusted items =3+5=8Rusted nails =3Total nails = 6Pgetting a rusted item or a nail=Pgetting a rusted item+Pgetting a nail-Pgetting a rusted item and a nail=816+616-316=8+6-316=1116

Page No 31.105:

Question 19:

A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the same colour is
(a) 5108

(b) 18108

(c) 30108

(d) 48108

Answer:

d 48108Psame coloured socks=Pboth brown+Pboth white=59×48+49×38=2072+1272=3272=49=48108

Page No 31.105:

Question 20:

If S is the sample space and P (A) = 13 P (B) and S = AB, where A and B are two mutually exclusive events, then P (A) =
(a) 1/4
(b) 1/2
(c) 3/4
(d) 3/8

Answer:

a 14PA=13PBPB=3PA           ...1A and B are mutually exclusive events.⇒ PAB=0Now,PAB=PA+PB=PSPA+PB=1PA+3PA=1           From 14PA=1PA=14

Page No 31.105:

Question 21:

If A and B are two events, then P (AB) =
(a) P A P B
(b) 1 − P (A) − P (B)
(c) P (A) + P (B) − P (AB)
(d) P (B) − P (AB)

Answer:

(d) P (B) − P (AB)



From the diagram, we get AB and A¯ B are mutually exclusive events such that (AB) (A¯B)=B. Therefore by addition theorem of probability we have P(AB) + P(A¯B) = P(B) PAB=PB-PAB

Page No 31.105:

Question 22:

If P (AB) = 0.8 and P (AB) = 0.3, then P A + P B =
(a) 0.3
(b) 0.5
(c) 0.7
(d) 0.9

Answer:

d 0.9PAB=PA+PB-PABPA+PB=PAB+PABPA+PB=0.8 +0.3PA+PB=1.11-PA¯+1-PB¯=1.1PA+PB=2-1.1PA+PB=0.9

Page No 31.105:

Question 23:

A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is
(a) 2/15
(b) 7/15
(c) 8/15
(d) 14/15

Answer:

(c) 8/15

A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag X and then drawing a white ball from it.
(II) Selecting bag Y and then drawing a white ball from it.

Let E1, E2 and A be three events as defined below:
E1 = Selecting bag X
E2 = Selecting bag Y
A = Drawing a white ball

We know that one bag is selected randomly.

 PE1=12      PE2=12 PA/E1=25PA/E2=46=23Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =12×25+12×23                                                   =15+13                                                   =3+515=815

Page No 31.105:

Question 24:

Two persons A and B take turns in throwing a pair of dice. The first person to throw 9 from both dice will be awarded the prize. If A throws first, then the probability that B wins the game is
(a) 9/17
(b) 8/17
(c) 8/9
(d) 1/9

Answer:

b  8179 can be obtained from throw of two dice in only 4 cases as given below:3, 64, 55, 46, 3⇒Pgetting 9 =436=19    Pnot getting 9=3236=89Now,PB is winning = Pgetting 9 in 2nd throw + Pgetting 9 in 4th throw+Pgetting 9 in 6th throw + ...=89×19+89×89×89×19+ ... =8811+6481+64812+ ...=881×11-6481=881×8117=817

Page No 31.105:

Question 25:

The probability that in a year of 22nd century chosen at random, there will be 53 Sunday, is
(a) 3/28
(b) 2/28
(c) 7/28
(d) 5/28

Answer:

​P(53 Sundays in a leap year) = 27

P(53 Sundays in a non-leap year) = 17

There will be 24 leap years in the 22nd century, i.e. from the year 2201 to 2200, we will have 24 leap years.

∴ P(leap year) = 24100

    P(non-leap year) = 76100

Now,

P(53 Sundays) = P(leap year)×P(53 Sundays in a leap year)
                                        + P(non-leap year)×P(53 Sundays in a non-leap year)

                                              =24100×27+76100×17=48700+76700=124700=31175


Disclaimer: None of the given options is correct.

Page No 31.105:

Question 26:

From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24 is
(a) 6/25
(b) 1/4
(c) 1/6
(d) 2/5

Answer:

Number divisible by 6 between 1 to 100 = 16
Number divisible by 8 between 1 to 100 = 12
Number divisible by 6 and 8 between 1 to 100 = 4
Number divisible by 24 between 1 to 100 = 4

P(number divisible by 6 or 8) = P(number divisible by 6) + P(number divisible by 8) - P(number divisible by 6 and 8)

                                               =16100+12100-4100=24100=625


P(number divisible by 6 or 8 but not by 24) = P(number divisible by 6 or 8) - P(number divisible by 24)

                                                                       =625-4100=625-125=525=15

Disclaimer: None of the given options is correct.

Page No 31.105:

Question 27:

Mark the correct alternative in the following question:

If A and B are two events such that P(A) = 45, and PAB=710, then P(B|A) =

a 110                                              b 18                                              c 78                                              d 1720

Answer:

We have,PA=45 and PAB=710Now,PB|A=PABPA=71045=7×510×4=78

Hence, the correct alternative is option (c).

Page No 31.105:

Question 28:

Choose the correct alternative in the following question:

If A and B are two events associated to a random experiment such that PAB=710 and PB=1720, then P(A|B) =

a 1417                                             b 1720                                             c 78                                             d 18

Answer:

We have,PAB=710 and PB=1720Now,PA|B=PABPB=7101720=7×2017×10=1417

Hence, the correct alternative is option (a).



Page No 31.106:

Question 29:

Choose the correct alternative in the following question:

Associated to a random experiment two events A and B are such that PB=35, PA|B=12 and PAB=45. The value of P(A) is

a 310                                             b 12                                             c 110                                             d 35

Answer:

We have,PB=35, PA|B=12 and PAB=45As, PA|B=12PABPB=12PAB=12×PBPAB=12×35PAB=310Now,PAB=45PA+PB-PAB=45PA+35-310=45PA+6-310=45PA+310=45PA=45-310PA=8-310PA=510 PA=12

Hence, the correct alternative is option (b).

Page No 31.106:

Question 30:

Choose the correct alternative in the following question:

If PA=310, PB=25 and PAB=35, then PA|B+PB|A equalsa 14                                        b 712                                        c 512                                        d 13

Answer:

We have,PA=310, PB=25 and PAB=35As, PAB=35PA+PB-PAB=35310+25-PAB=353+410-PAB=35PAB=710-35PAB=7-610PAB=110Now,PA|B+PB|A=PABPB+PABPA=11025+110310=510×2+1010×3=14+13=3+412=712

Hence, the correct alternative is option (b).

Disclaimer: The option (b) given in the book is incorrect as the probability of any event is always less than 1. The same has been corrected here.

 

Page No 31.106:

Question 31:

Choose the correct alternative in the following question:

Let PA=713, PB=913 and PAB=413. Then, PA|B=a 59                                        b 49                                        c 413                                        d 613

Answer:

We have,PA=713, PB=913 and PAB=413As, PAB=PB-PAB=913-413=513Now,PA|B=PABPB=513913=59

Hence, the correct alternative is option (a).

Page No 31.106:

Question 32:

Choose the correct alternative in the following question:

If PA=25, PB=310 and PAB=15, then, PA|B PB|A is equal toa 56                                        b 57                                        c 2542                                        d 1

Answer:

We have,PA=25, PB=310 and PAB=15Also, PA=1-PA=1-25=5-25=35 andPB=1-PB=1-310=10-310=710As, PAB=PA+PB-PAB=25+310-15=4+3-210=510=12Also, PAB=PAB=1-PAB=1-12=12Now,PA|B×PB|A=PABPB×PABPA=12710×1235=102×7×53×2=57×56=2542

Hence, the correct alternative is option (c).

Page No 31.106:

Question 33:

Mark the correct alternative in the following question:

If A and B are two events such that PA=12, PB=13, PA|B=14, then PAB equalsa 112                                            b 34                                            c 14                                            d 316

Answer:

We have,PA=12, PB=13 and PA|B=14As, PA|B=14PABPB=14PAB=14×PBPAB=14×13PAB=112Also, PAB=PA+PB-PAB=12+13-112=6+4-112=912=34Now,PAB=PAB=1-PAB=1-34=4-34=14

Hence, the correct alternative is option (c).

Page No 31.106:

Question 34:

Mark the correct alternative in the following question:

Let A and B are two events such that PA=38, PB=58 and PAB=34. Then PA|B×PAB is equals toa 25                                            b 38                                            c 320                                            d 625

Answer:

We have,PA=38, PB=58 and PAB=34As, PAB=34PA+PB-PAB=3438+58-PAB=3488-PAB=341-PAB=34PAB=1-34PAB=14Also,PAB=PB-PAB=58-14=5-28=38Now,PA|B×PA|B=PABPB×PABPB=1458×3858=84×5×3×85×8=25×35=625

Hence, the correct alternative is option (d).

Page No 31.106:

Question 35:

Mark the correct alternative in the following question:

If PB=35, PA|B=12 and PAB=45, then PAB+PAB=a 15                                            b 45                                            c 12                                            d 1

Answer:

We have,PB=35, PA|B=12 and PAB=45As, PA|B=12PABPB=12PAB=12×PBPAB=12×35PAB=310As, PAB=45PA+PB-PAB=45PA+35-310=45PA+310=45PA=45-310PA=510PA=12Now,PAB+PAB=1-PAB+1-POnly A=1-45+1-PA-PAB=15+1-12-310=65-210=1010=1

Hence, the correct alternative is option (d).

Page No 31.106:

Question 36:

Mark the correct alternative in the following question:

If PA=0.4, PB=0.8 and PB|A=0.6, then PAB=a 0.24                                      b 0.3                                      c 0.48                                      d 0.96

Answer:

We have,PA=0.4, PB=0.8 and PB|A=0.6As, PB|A=0.6PABPA=0.6PAB=0.6×PAPAB=0.6×0.4PAB=0.24Now, PAB=PA+PB-PAB=0.4+0.8-0.24=1.2-0.24=0.96

Hence, the correct alternative is option (d).

Page No 31.106:

Question 37:

Mark the correct alternative in the following question:

If PB=35, PA|B=12 and PAB=45, then PB|A=a 15                                      b 310                                      c 12                                      d 35

Answer:

We have,PB=35, PA|B=12 and PAB=45As, PA|B=12PABPB=12PAB=12×PBPAB=12×35PAB=310And, PBA=PB-PAB=35-310=310Also, PAB=PA+PB-PAB45=PA+35-31045=PA+310PA=45-310PA=510PA=12So, PA=1-PA=1-12=12Now,PB|A=PBAPA=31012=3×210=35

Hence, the correct alternative is option (d).

Page No 31.106:

Question 38:

Mark the correct alternative in the following question:

If A and B are two events such that PA=0.4, PB=0.3 and PAB=0.5, then PBA equalsa 23                                      b 12                                      c 310                                      d 15

Answer:

We have,PA=0.4, PB=0.3 and PAB=0.5As, PAB=0.5PA+PB-PAB=0.50.4+0.3-PAB=0.50.7-PAB=0.5PAB=0.7-0.5PAB=0.2Now,PBA=PA-PAB=0.4-0.2=0.2=210=15

Hence, the correct alternative is option (d).

Page No 31.106:

Question 39:

If A and B are two events such that A ≠ Φ, B = Φ, then

(a) PAB=PABPB             (b) PAB=PA PB
(c) PAB=PBA=1           (d) PAB=PAPB 

Answer:

By the definition of conditional probability:
 If A and B are two events such that A ≠ Φ, B = Φ, then PAB=PABPB.
Hence, the correct answer is option (a).

Page No 31.106:

Question 40:

Mark the correct alternative in the following question:

If A and B are two events such that PA0 and PB1, then PA|B=a 1-PA|B                                   b 1-PA|B                                   c 1-PABPB                                   d PAPB

Answer:

We have,PA0 and PB1Now,PA|B=PABPB=PABPB=1-PABPB

Hence, the correct alternative is option (c).

Page No 31.106:

Question 41:

Mark the correct alternative in the following question:

If the events A and B are independent, then PAB is equal toa PA+PB                                   b PA-PB                                   c PA PB                                   d PAPB

Answer:

As, A and B are independent events.So, PA|B=PA and PB|A=PBNow,PA|B=PABPBPAB=PB PA|BPAB=PB PA

Hence, the correct alternative is option (c).



Page No 31.107:

Question 42:

Mark the correct alternative in the following question:

If A and B are two independent events with PA=35 and PB=49, then PAB equalsa 415                                       b 845                                       c 13                                       d 29

Answer:

We have,PA=35 and PB=49As, A and B are independent eventsSo, PAB=PA×PB=35×49=1245=415Also, PAB=PA+PB-PAB=35+49-415=27+20-1245=3545=79Now,PAB=PAB=1-PAB=1-79=29

Hence, the correct alternative is option (d).

Page No 31.107:

Question 43:

Mark the correct alternative in the following question:

If A and B are two independent events such that PA=0.3 and PAB=0.5, then PA|B-PB|A=a 27                                              b 335                                             c 170                                             d 17

Answer:

We have,PA=0.3 and PAB=0.5As, A and B are independent eventsSo, PAB=PA×PB=0.3×PB=0.3PB                     .....iAlso, PAB=PA+PB-PAB0.5=0.3+PB-0.3PB                      Using i0.5-0.3=0.7PB0.7PB=0.2PB=0.20.7PB=27Using i, we getPAB=0.3×27=670Now,PA|B-PB|A=PABPB-PABPA=67027-6700.3=6×770×2-670×0.3=310-27=21-2070=170

Hence, the correct alternative is option (c).

Page No 31.107:

Question 44:

Mark the correct alternative in the following question:

A flash light has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, then the probability that both are dead is

a 328                                               b 114                                               c 964                                               d 3356

Answer:

We have,The total number of batteries=8The number of dead batteries=3Let A be the event of selecting the first dead battery andB be the event of selecting the second dead batteryNow,Pboth dead batteries are selected=PAB=PA×PB|A=38×27=328

Hence, the correct alternative is option (a).

Page No 31.107:

Question 45:

Mark the correct alternative in the following question:

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability of getting exactly one red ball is

a 1529                                   b 1556                                   c 45196                                   d 135392

Answer:

We have,The number of red balls=5 andThe number of blue balls=3Let R be the event of getting a red ball andB be the event of getting a blue ball.Now,PGetting exactly one red ball=PRBB+PBRB+PBBR=PR×PB|R×PB|RB+PB×PR|B×PB|BR+PB×PB|B×PR|BB=58×37×26+38×57×26+38×27×56=556+556+556=1556

Hence, the correct alternative is option (b).

Page No 31.107:

Question 46:

Mark the correct alternative in the following question:

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability that exactly two of the three balls were red, the first ball being red, is

a 13                                             b 47                                             c 1528                                             d 528

Answer:

We have,The number of red balls=5 andThe number of blue balls=3Let R be the event of getting a red ball andB be the event of getting a blue ball.Now,PGetting exactly two red balls of the three balls, the first ball being red=PRB|R+PBR|R=47×36+37×46=27+27=47

Hence, the correct alternative is option (b).

Page No 31.107:

Question 47:

Mark the correct alternative in the following question:

In a college 30% students fail in Physics, 25% fail in Mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in Physics if she failed in Mathematics is

a 110                                      b 13                                      c 25                                      d 920

Answer:

Let A be the event of choosing a student failed in Physics andB be the event of choosing a student failed in Mathematics.We have,PA=30%=30100=310,PB=25%=25100=14 andPAB=10%=10100=110Now,PA|B=PABPB=11014=410=25

Hence, the correct alternative is option (c).

Page No 31.107:

Question 48:

Mark the correct alternative in the following question

Three persons, A, B and C fire a target in turn starting with A. Their probabilities of hitting the target are 0.4, 0.2 and 0.2, respectively. The probability of two hits is

(a) 0.024                                             (b) 0.452                                             (c) 0.336                                             (d) 0.188

Answer:

Let:A be the event of hitting the target by the person A,B be the event of hitting the target by the person B andC be the event of hitting the target by the person CWe have,PA=0.4, PB=0.3 and PC=0.2Also,PA=1-PA=1-0.4=0.6,PB=1-0.3=0.7 andPC=1-0.2=0.8Now,PTwo hits=PABC+PABC+PABC=PA×PB×PC+PA×PB×PC+PA×PB×PC=0.4×0.3×0.8+0.4×0.7×0.2+0.6×0.3×0.2=0.096+0.056+0.036=0.188

Hence, the correct alternative is option (d).

Disclaimer: The option (d) in the textbook is incorrect. It should be 0.188 instead 0.138. The same has been corrected here.

Page No 31.107:

Question 49:

A and B are two students. Their chances of solving a problem correctly are 13 and 14 respectively. If the probability of their making common error is 120 and they obtain the same answer, then the probability of their answer to be correct is
(a) 1013            (b) 13120             (c) 140             (d) 112

Answer:

E1 = they solve correctly.
E2 = they solve incorrectly.
A = they obtain the same result.
PE1A=PE1 PAE1PE1 PAE1+PE2 PAE2                =112×1112×1+612×120                =112×1112×1+612×120                =2026=1013
Hence, the correct answer is option (a).

Page No 31.107:

Question 50:

Mark the correct alternative in the following question:

Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability that both cards are queen is

a 113×113                                     b 113+113                                     c 113×117                                     d 113×45

Answer:

Let:A be the event that a queen is drawn in the first draw andB be the event that a queen is drawn in the second draw as wellNow,PBoth the two cards drawn are queen=PA×PB|A=452×452=113×113

Hence, the correct alternative is option (a).

Page No 31.107:

Question 51:

Mark the correct alternative in the following question:

A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

a 167168                                             b 128                                             c 221                                             d 328

Answer:

Let:O be the event of drawing a orange ball,G be the event of drawing a green ball andB be the event of drawing a blue ballWe have,nO=3, nG=3 and nB=2Also, total balls=8Now,Pdrawing 2 green balls and one blue ball=PGGB+PGBG+PBGG=PG×PG|G×PB|GG+PG×PB|G×PG|GB+PB×PG|B×PG|BG=38×27×26+38×27×26+28×37×26=128+128+128=328

Hence, the correct alternative is option (d).

Page No 31.107:

Question 52:

Mark the correct alternative in the following question:

If two events are independent, then

(a) they must be mutually exclusive
(b) the sum of their probabilities must be equal to 1
(c) (a) and (b) both are correct
(d) none of the above is correct

Answer:

Let A and B are two independent events. Then,PAB=PA×PBAs, PAB0 or PA+PB1So, both are neither mutually exclisive nor their sum of probabilities is 1.

Hence, the correct alternative is option (d).



Page No 31.108:

Question 53:

Mark the correct alternative in the following question:

Two dice are thrown. If it is known that the sum of the numbers on the dice was less than 6, then the probability of getting a sum 3, is

a 118                                           b 518                                           c 15                                           d 25

Answer:

Let:A be the event of getting a sum of 3 andB be the event of getting a sum of 6As, A=1,2,2,1 and B=1,1,1,2,1,3,1,4,2,1,2,2,2,3,3,1,3,2,4,1So, nA=2, nB=10 and nAB=nA=2Now,PA|B=nABnB=210=15

Hence, the correct alternative is option (c).

Page No 31.108:

Question 54:

Mark the correct alternative in the following question:

If A and B are such that PAB=59 and PAB=23, then PA+PB=a 910                                           b 109                                           c 89                                           d 98

Answer:

We have,PAB=59 and PAB=23As, PAB=23PAB=23PAB=1-23PAB=13Also,PAB=PA+PB-PABPA+PB=PAB+PAB=59+13PA+PB=89       .....iNow,PA+PB=1-PA+1-PB=2-PA+PB=2-89                        Using i=109

Hence, the correct alternative is option (b).

Page No 31.108:

Question 55:

Mark the correct alternative in the following question:

If A and B are two events such that PA|B=p, PA=p, PB=13 and PAB=59, then p=a 23                                            b 35                                            c 13                                            d 34

Answer:

We have,PA|B=p, PA=p, PB=13 and PAB=59As, PA|B=pPABPB=pPAB=p×PBPAB=p×13PAB=p3Now,PAB=59PA+PB-PAB=59p+13-p3=59p-p3=59-132p3=29p=29×32 p=13

Hence, the correct alternative is option (c).

Page No 31.108:

Question 56:

Mark the correct alternative in the following question:

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number of the die and a spade card is

a 12                                                        b 14                                                        c 18                                                        d 34

Answer:

Let:A be the event of getting an even number of the die andB be the event of getting a spade cardNow,Pgetting an even number of the die and a spade card=PAB=PA×PB|A=36×1352=18

Hence, the correct alternative is option (c).

Page No 31.108:

Question 57:

Mark the correct alternative in the following question:

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is

a 12                                                        b 13                                                        c 23                                                        d 47

Answer:

We have,S=BBB,BBG,BGB,BGG,GGG,GBG,GGB,GBB, where the first letter in each element represents the eldest childLet:A be the event of choosing a family with a girl as the eldest child andB be the event of choosing a family with at least one girl childSo, A=GGG,GBG,GGB,GBB and B=BBG,BGB,BGG,GGG,GBG,GGB,GBBnA=4, nB=7 and nAB=nA=4Now,PA|B=nABnB=47

Hence, the correct alternative is option (d).

Page No 31.108:

Question 58:

Mark the correct alternative in the following question:

Let A and B be two events. If PA=0.2, PB=0.4, PAB=0.6, then PA|B is equal toa 0.8                                                 b 0.5                                                 c 0.3                                                 d 0

Answer:

We have,PA=0.2, PB=0.4 and PAB=0.6As, PAB=0.6PA+PB-PAB=0.60.2+0.4-PAB=0.60.6-PAB=0.6PAB=0.6-0.6PAB=0Now,PA|B=PABPB=00.4=0

Hence, the correct alternative is option (d).

Page No 31.108:

Question 59:

Mark the correct alternative in the following question:

Let A and B be two events such that PA=0.6, PB=0.2, PA|B=0.5. Then PA|B equalsa 110                                                 b 310                                                 c 38                                                 d 67

Answer:

We have,PA=0.6, PB=0.2, PA|B=0.5As, PA|B=0.5PABPB=0.5PAB=0.5×PBPAB=0.5×0.2PAB=0.1Also, PB=1-PB=1-0.2=0.8As, PAB=PA+PB-PAB =0.6+0.2-0.1=0.7Now,PA|B=PABPB=PAB0.8=1-PAB0.8=1-0.70.8=0.30.8=38

Hence, the correct alternative is option (c).

Disclaimer: The answer given in the book is incorrect. The same has been corrected here.



Page No 31.17:

Question 1:

Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Answer:

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Consider the given events.
A = An even number on the card
B = A number more than 3 on the card

Clearly,
A = {2, 4, 6, 8, 10}
B = {4, 5, 6, 7, 8, 9, 10}

Now, AB=4,6,8,10 Required probability =  PA/B=nABnB=47

Page No 31.17:

Question 2:

Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Answer:

Consider the given events.
A = Both the children are girls.
B = The youngest child is a girl.
C = At least one child is a girl.

Clearly, S=B1B2, B1G2, G1B2, G1G2A=G1G2B=B1G2, G1G2 C=B1G2,G1B2,G1G2

AB=G1G2 and AC=G1G2i Required probability =  PA/B=nABnB=12ii Required probability =  PA/C=nABnC=13

Page No 31.17:

Question 3:

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.

Answer:

Consider the given events
A = Numbers appearing on two dice are different
B = The sum of the numbers on two dice is 4

Clearly,
A = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
                     (2, 1), (2, 3), (2, 4), (2, 5), (2, 6),
                     (3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
                     (4, 1), (4, 2), (4, 3), (4, 5), (4, 6),
                     (5, 1), (5, 2), (5, 3), (5, 4), (5, 6),
                     (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
B = {(1, 3), (3, 1) and (2, 2)}

Now, AB=1,3 and 3,1 Required probability =  PB/A=nABnA=230=115

Page No 31.17:

Question 4:

A coin is tossed three times, if head occurs on first two tosses, find the probability of getting head on third toss.

Answer:

Consider the given events.
A = Getting head on third toss
B = Getting head on first two tosses

Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}

Now,AB=H,H,H Required probability =  PA/B=nABnB=12

Page No 31.17:

Question 5:

A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.

Answer:

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and 5 on second throw

Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4),
                     (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4),
                    .                                                                                 . 
                    .                                                                                 . 
                    .                                                                                 .                                                                          
                    (6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

Now, AB=6,5,4 Required probability =  PA/B=nABnB=16

Page No 31.17:

Question 6:

Compute P (A/B), if P (B) = 0.5 and P (AB) = 0.32

Answer:

Given:PB=0.5PAB=0.32Now, PAB=PABPB PAB=0.320.5=3250=1625

Page No 31.17:

Question 7:

If P (A) = 0.4, P (B) = 0.3 and P (B/A) = 0.5, find P (AB) and P (A/B).

Answer:

Given:PA=0.4PB=0.3PB/A=0.5Now,PB/A=PABPA0.5=PAB0.4PAB=0.2PA/B=PABPB=0.20.3=23

Page No 31.17:

Question 8:

If A and B are two events such that P (A) = 13, P (B) = 15 and P (AB) = 1130, find P (A/B) and P (B/A).

Answer:

Given:PA=13PB=15PAB=1130Now, PAB=PA+PB-PAB1130=13+15-PABPAB=13+15-1130=10+6-1130=530=16PA/B=PABPB=1615=56PB/A=PABPA=1613=36=12

Page No 31.17:

Question 9:

A couple has two children. Find the probability that both the children are (i) males, if it is known that at least one of the children is male. (ii) females, if it is known that the elder child is a female.

Answer:

Consider the given events.
A = Both the children are female.
B = The elder child is a female.
C = At least one child is a male.
D = Both children are male.

Clearly, S=M1M2, M1F2, F1M2, F1F2A=F1F2B=F1M2, F1F2C=M1F2, F1M2, M1M2 D=M1M2
                                     [Here, first child is elder and second is younger]

DC=M1M2 and AB=F1F2i Required probability =  PD/C=nDCnC=13ii Required probability =  PA/B=nABnB=12



Page No 31.22:

Question 1:

From a pack of 52 cards, two are drawn one by one without replacement. Find the probability that both of them are kings.

Answer:

Consider the given events.
A = A king in the first draw
B = A king in the second draw

Now, PA=452=113PB/A= Getting a king in the second draw after getting a king in the first draw            =351                 After the first draw, the total number of cards will be 51. Then, 3 kings will be remaining.            =117 Required probability =  PAB=PA×PB/A=113×117=1221

Page No 31.22:

Question 2:

From a pack of 52 cards, 4 are drawn one by one without replacement. Find the probability that all are aces(or kings).

Answer:

Consider the given events.
A = An ace in the first draw
B = An ace in the second draw
C = An ace in the third draw
D = An ace in the fourth draw

Now, PA=452=113PB/A=351=117PC/AB=250=125PD/ABC=149 Required probability =  PABCD=PA×PB/A×PC/AB×PD/ABC                                                                            =113×117×125×149                                                                            =1270725

In case of kings, the required probablity will be = 1270725 

Page No 31.22:

Question 3:

Find the chance of drawing 2 white balls in succession from a bag containing 5 red and 7 white balls, the ball first drawn not being replaced.

Answer:

Consider the given events.
A = A white ball in the first draw
B = A white ball in the second draw

Now, PA=712PB/A=611 Required probability =  PAB=PA×PB/A=712×611=722

Page No 31.22:

Question 4:

A bag contains 25 tickets, numbered from 1 to 25. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show even numbers.

Answer:

There are 12 even numbers between 1 to 25.

Consider the given events.
A = An even number ticket in the first draw
B = An even number ticket in the second draw

Now, PA=1225PB/A=1124 Required probability =  PAB=PA×PB/A=1225×1124=1150

Page No 31.22:

Question 5:

From a deck of cards, three cards are drawn on by one without replacement. Find the probability that each time it is a card of spade.

Answer:

Consider the events
A = An ace in the first draw
B = An ace in the second draw
C = Getting an ace in the third draw

Now, PA=1352=14PB/A=1251=417PC/AB=1150 Required probability =  PABC                                 =PA×PB/A×PC/AB                                 =14×417×1150                                 =11850

Page No 31.22:

Question 6:

Two cards are drawn without replacement from a pack of 52 cards. Find the probability that
(i) both are kings
(ii) the first is a king and the second is an ace
(iii) the first is a heart and second is red.

Answer:

(i) Consider the given events
A = A king in the first draw
B = A king in the second draw

Now, PA=452=113PB/A=351=117 Required probability =  PAB                                 =PA×PB/A                                 =113×117                                 =1221

(ii) Consider the given events
A = A king in the first draw
B = An ace in the second draw

Now, PA=452=113PB/A=451=451 Required probability =  PAB                                 =PA×PB/A                                 =113×1451                                 =4663

(iii) Consider the given events.
A = A heart in the first throw
B = A red card in the second throw

Now,PA=1352=14PB/A=2551 Required probability =  PAB                                 =PA×PB/A                                 =14×2551                                 =25204
 

Page No 31.22:

Question 7:

A bag contains 20 tickets, numbered from 1 to 20. Two tickets are drawn without replacement. What is the probability that the first ticket has an even number and the second an odd number.

Answer:

There are 10 even numbers and 10 odd numbers between 1 to 20.

Consider the given events.
A = An even number in the first draw
B = An odd number in the second draw

Now, PA=1020=12PB/A=1019 Required probability=PAB=PA×PB/A=12×1019=519

Page No 31.22:

Question 8:

An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?

Answer:

Consider the given events.
A = A white or red ball in the first draw
B = A white or red ball in the second draw

Now, PA=712PB/A=611 PAB=PA×PB/A                  =712×611                  =722 Required probability =  1-PAB                                  =1-722                                  =1522
 

Page No 31.22:

Question 9:

A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red.

Answer:

Consider the given events.
A = A white or black ball in the first draw
B = A white or black ball in the second draw
C = A white or black ball in the third draw

Now, PA=815PB/A=714=12PC/AB=613 Required probability =  PABC                                 =PA×PB/A×PC/AB                                 =815×12×613                                 =865

Page No 31.22:

Question 10:

A card is drawn from a well-shuffled deck of 52 cards and then a second card is drawn. Find the probability that the first card is a heart and the second card is a diamond if the first card is not replaced.

Answer:

Consider the given events.
A = A heart in the first draw
B = A diamond in the second draw

Now, PA=1352=14PB/A=1351 Required probability =  PAB=PA×PB/A=14×1351=13204

Page No 31.22:

Question 11:

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Answer:

Consider the given events.
A = A black ball in the first draw
B = A black ball in the second draw

Now, PA=1015=23PB/A=914 Required probability =  PAB=PA×PB/A=23×914=37

Page No 31.22:

Question 12:

Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and third card drawn is an ace?

Answer:

Consider the given events.
A = A king in the first draw
B = A king in the second draw
C = An ace in the third draw


Now,PA=452=113PB/A=351=117PC/AB=450=225 Required probability =  PABC                                 =PA×PB/A×PC/AB                                 =113×117×225                                 =25525

Page No 31.22:

Question 13:

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Answer:

Consider the given events.
A = A good orange in the first draw
B = A good orange in the second draw
C = A good orange in the third draw


Now, PA=1215=45PB/A=1114PC/AB=1013 Required probability =  PABCD                                =PA×PB/A×PC/AB                                =45×1114×1013                                =4491

Page No 31.22:

Question 14:

A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.

Answer:

Consider the given events.
A = A white ball in the first draw
B = A black ball in the second draw
C = A red ball in the third draw

Now, PA=416=14PB/A=715PC/AB=514 Required probability =  PABC=PA×PB/A×PC/AB                                                          =14×715×514                                                          =124



Page No 31.34:

Question 1:

If P (A) = 713, P (B) = 913 and P (AB) = 413, find P (A/B).

Answer:

Given:PA=713PB=913 PAB=413Now,PAB=PABPB PAB=413913=49

Page No 31.34:

Question 2:

If A and B are events such that P (A) = 0.6, P (B) = 0.3 and P (AB) = 0.2, find P (A/B) and P (B/A).

Answer:

Given:PA=0.6PB=0.3 PAB=0.2Now,PAB=PABPB PAB=0.20.3=23PBA=PABPA PBA=0.20.6=13

Page No 31.34:

Question 3:

If A and B are two events such that P (AB) = 0.32 and P (B) = 0.5, find P (A/B).

Answer:

Given:PB=0.5 PAB=0.32Now,PAB=PABPB PAB=0.320.5=3250=0.64

Page No 31.34:

Question 4:

If P (A) = 0.4, P (B) = 0.8, P (B/A) = 0.6. Find P (A/B) and P (AB).

Answer:

Given:PA=0.4PB=0.8 PB/A=0.6Now,PB/A=PABPA0.6=PAB0.4PAB=0.24PA/B=PABPB=0.240.8=0.3PAB=PA+PB-PABPAB=0.4+0.8-0.24=0.96

Page No 31.34:

Question 5:

If A and B are two events such that

i PA=13, PB=14 and PAB=512, then find PA|B and PB|A.ii PA=611, PB=511 and PAB=711, then find PAB, PA|B and PB|A.iii PA=713, PB=913 and PAB=413, then find PA|B.iv PA=12, PB=13 and PAB=14, then find PA|B, PB|A, PA|B and PA|B.
                                                                                                                                                                                         [NCERT EXEMPLAR]

Answer:

i We have,PA=13, PB=14 and PAB=512As, PAB=PA+PB-PABPAB=PA+PB-PABPAB=13+14-512PAB=212PAB=16Now,PA|B=PABPB=1614=46=23 andPB|A=PABPA=1613=36=12ii We have,PA=611, PB=511 and PAB=711As, PAB=PA+PB-PABPAB=PA+PB-PABPAB=611+511-711PAB=6+5-711PAB=411Now,PA|B=PABPB=411511=45 andPB|A=PABPA=411611=46=23iii We have,PA=713, PB=913 and PAB=413As, PAB=PB-PABPAB=913-413PAB=9-413PAB=513Now,PA|B=PABPB=513913=59iv We have,PA=12, PB=13 and PAB=14Also, PB=1-PB=1-13=23As, PAB=PA+PB-PAB=12+13-14=6+4-312PAB=712Also, PAB=PB-PABPAB=13-14PAB=4-312PAB=112And, PAB=PAB=1-PAB=1-712=512Now,PA|B=PABPB=1413=34,PB|A=PABPA=1412=24=12,PA|B=PABPB=11213=312=14 andPA|B=PABPB=51223=1524=58

Page No 31.34:

Question 6:

If A and B are two events such that 2 P (A) = P (B) = 513 and P (A/B) = 25, find P (AB).

Answer:

Given: 2PA=PB=513 PA/B=25 PA=526      PB=513Now, PA/B=PABPB25=PAB513PAB=25×513=213PAB=PA+PB-PAB                     =526+513-213                     =1126

Page No 31.34:

Question 7:

If P (A) = 611, P (B) = 511 and P (AB) = 711, find
(i) P (AB)
(ii) P (A/B)
(iii) P (B/A)

Answer:

Given: PA=611PB=511 PAB=711(i) PAB=PA+PB-PAB711=611+511-PABPAB=611+511-711=411ii PA/B=PABPB                    =411511                    =45iii PB/A=PABPA                     =411611                      =46                      =23

Page No 31.34:

Question 8:

A coin is tossed three times. Find P (A/B) in each of the following:
(i) A = Heads on third toss, B = Heads on first two tosses
(ii) A = At least two heads, B = At most two heads
(iii) A = At most two tails, B = At least one tail.

Answer:

(i) Consider the given events.
A = Heads on third toss
B = Heads on first two tosses

Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}

Now, AB=H, H, H Required probability=PA/B=nABnB=12

(ii) Consider the given events.
A = At least two heads
B = At most two heads

Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (H, H, T)}
B = {(T, T, T), (H, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T)}

Now, AB=H, T, H, T, H, H, H, H, T Required probability =  PA/B=nABnB=37

(iii) Consider the given events.
A = At most two tails
B = At least one tail

Clearly,
A = {(T, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T), (H, H, H)}
B = {(T, T, T), (T, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T)}

Now, AB={(T, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T)} 

 Required probability=PA/B=nABnB=67

Page No 31.34:

Question 9:

Two coins are tossed once. Find P (A/B) in each of the following:
(i) A = Tail appears on one coin, B = One coin shows head.
(ii) A = No tail appears, B = No head appears.

Answer:

(i) Consider the given events.
A = Tail appears on one coin
B = One coin shows head

Clearly,
A = {(H, T), (T, H)}
B = {(H, T), (T, H)}

Now, AB=H, T, T, H Required probability=PA/B=nABnB=22=1

(ii) Consider the given events.
A = No tail appears
B = No head appears

Clearly,
A = {(H, H)}
B = {(T, T)}

Now, AB=ϕ Required probability =  PA/B=nABnB=01=0

Page No 31.34:

Question 10:

A die is thrown three times. Find P (A/B) and P (B/A), if
A = 4 appears on the third toss, B = 6 and 5 appear respectively on first two tosses.

Answer:

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw

Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4),
                    .                                                                                 . 
                    .                                                                                 . 
                    .                                                                                 .                                                                          
                    (6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

Now, AB=6, 5, 4 Required probability =  PA/B=nABnB=16 Required probability =  PB/A=nABnA=136

Page No 31.34:

Question 11:

Mother, father and son line up at random for a family picture. If A and B are two events given by A = Son on one end, B = Father in the middle, find P (A/B) and P (B/A).

Answer:

Consider the given events.
A = Son standing on one end
B = Father standing in the middle

Clearly, S=MFS, MSF, FSM, FMS, SMF, SFMA=MFS, FMS, SMF, SFM, B=MFS, SFM

Now,AB=MFS, SFM i Required probability =  PA/B=nABnB=22=1ii Required probability =  PB/A=nABnA=24=12

Page No 31.34:

Question 12:

A dice is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Answer:

Consider the given events.
A = 4 appears on the die
B = The sum of the numbers on two dice is 6.

Clearly,
A = {(1, 4) (2, 4), (3, 4),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 4), (6, 4)}
B = {(1, 5), (5, 1), (2, 4), (4, 2),(3, 3)}

Now, AB=2,4 and 4,2 Required probability =  PA/B=nABnB=25

Page No 31.34:

Question 13:

Two dice are thrown. Find the probability that the numbers appeared has the sum 8, if it is known that the second die always exhibits 4.

Answer:

Consider the given events.
A = 4 appears on second die
B = The sum of the numbers on two dice is 8.

Clearly,
A = {(1, 4), (2, 4), (3, 4), (4, 4) (5, 4) (6, 4)}
B = {(4, 4), (3, 5), (5, 3) (2, 6), (6, 2)}

Now, AB=4,4  Required probability =  PB/A=nABnA=16



Page No 31.35:

Question 14:

A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.

Answer:

Consider the given events.
A = Number appearing on second die is odd
B = The sum of the numbers on two dice is 7.

Clearly,
A = {(1, 1), (1, 3),(1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 3), (5, 5),(6, 1), (6, 3), (6, 5)}
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}

Now, AB=2,5,4,3,6,1 Required probability =  PB/A=nABnA=318=16

Page No 31.35:

Question 15:

A pair of dice is thrown. Find the probability of getting 7 as the sum if it is known that the second die always exhibits a prime number.

Answer:

Consider the given events
A = A prime number appears on second die.
B = The sum of the numbers on two dice is 7.

Clearly,
A = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3)(2, 5), (3, 2), (3, 3), (3, 5) (4, 2), (4, 3), (4, 5),(5, 2), (5, 3), (5, 5), (6, 2), (6, 3),(6, 5)}
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}

Now, AB={(2, 5), (5, 2),(4, 3)}

 Required probability =  PB/A=nABnA=318=16

Page No 31.35:

Question 16:

A die is rolled. If the outcome is an odd number, what is the probability that it is prime?

Answer:

Consider the given events.
A = The number is odd
B = The number is prime

Clearly,
A = {1, 3, 5}
B = {2, 3,5}

Now, AB=3,5 Required probability =  PB/A=nABnA=23

Page No 31.35:

Question 17:

A pair of dice is thrown. Find the probability of getting the sum 8 or more, if 4 appears on the first die.

Answer:

Consider the given events.
A = 4 appears on first die
B = The sum of the numbers on two dice is 8 or more.

Clearly,
A = {(4, 1), (4, 2), (4, 3), (4, 4) (4, 5), (4, 6)}
n(A) = 6

B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6),(5, 3), (5, 4), (5, 5) (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(B) = 15

Now, AB=4, 4, 4, 5, 4, 6 Required probability =  PB/A=nABnA=36=12

Page No 31.35:

Question 18:

Find the probability that the sum of the numbers showing on two dice is 8, given that at least one die does not show five.

Answer:

Consider the given events.
A = At least one die does not show 5
B = The sum of the numbers on two dice is 8.

Clearly,
A = {(1, 1), (1, 2) (1, 3), (1, 4),(1, 6),(2, 1), (2, 2) (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3) (3, 4), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)}
B = {(2, 6), (3, 5), (4, 4), (5, 3),(6, 2)}

Now, AB=4, 4, 6, 2, 2, 6 Required probability=PB/A=nABnA=325

Page No 31.35:

Question 19:

Two numbers are selected at random from integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.

Answer:

Suppose O represents the event of getting two odd numbers and S represents the event of getting their sum as an even number.Now,PO/S=POS PS=5C29C24C2+5C29C2=5C24C2+5C2=1016=58                                                                         

Page No 31.35:

Question 20:

A die is thrown twice and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once?

Answer:

Consider the given events.
A = 5 appears on the die at least once
B = The sum of the numbers on two dice is 8.

Clearly,
A = {(1, 5),(2, 5),(3, 5),(4, 5),(5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
B = {(2, 6), (3, 5), (4, 4), (5, 3),(6, 2)}

Now, AB=3,5,5,3 Required probability =  PA/B=nABnB=25

Page No 31.35:

Question 21:

Two dice are thrown and it is known that the first die shows a 6. Find the probability that the sum of the numbers showing on two dice is 7.

Answer:

Consider the given events.
A = First die shows 6
B = The sum of the numbers on two dice is 7.

Clearly,
A= {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(2, 5), (5, 2), (4, 3), (3, 4), (1, 6), (6, 1)}

Now, AB=6, 1 Required probability =  PB/A=nABnA=16

Page No 31.35:

Question 22:

A pair of dice is thrown. Let E be the event that the sum is greater than or equal to 10 and F be the event "5 appears on the first-die". Find P (E/F). If F is the event "5 appears on at least one die", find P (E/F).

Answer:

Consider the given events.
E = The sum of the numbers on two dice is 10 or more
F = 5 appears on first die

Clearly,
E = {(4, 6),(5, 5),(5, 6),(6, 4), (6, 5), (6, 6)}
F = {(5, 1), (5, 2), (5, 3), (5, 4) (5, 5), (5, 6)}

Now, EF=5, 5, 5, 6 Required probability =  PE/F=nEFnF=26=13

Second case:
Consider the given events.
E = The sum of the numbers on two dice is 10 or more
F = 5 appears on a die at least once

Clearly,
E = {(4, 6),(5, 5),(5, 6),(6, 4), (6, 5), (6, 6)}
F = {(1, 5),(2, 5),(3, 5),(4, 5),(5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}

Now, EF=5, 5, 5, 6, 6, 5 Required probability =  PE/F=nEFnF=311

Page No 31.35:

Question 23:

The probability that a student selected at random from a class will pass in Mathematics is 45, and the probability that he/she passes in Mathematics and Computer Science is 12. What is the probability that he/she will pass in Computer Science if it is known that he/she has passed in Mathematics?

Answer:

Consider the given events.
M = Students passes Mathematics
C = Students passes Computer Science

We have, PM=45 PMC=12Now,PCM=PMCPM              =1245=58

Page No 31.35:

Question 24:

The probability that a certain person will buy a shirt is 0.2, the probability that he will buy a trouser is 0.3, and the probability that he will buy a shirt given that he buys a trouser is 0.4. Find the probability that he will buy both a shirt and a trouser. Find also the probability that he will buy a trouser given that he buys a shirt.

Answer:

Suppose S represents the event of buying a shirt and T represents the event of buying a trouser.We have, PS=0.2PT=0.3 PS/T=0.4Now,PS/T=PST PTPST= PS/T× PT=0.4×0.3=0.12PT/S=PST PS=0.120.2=0.6                                                                         

Page No 31.35:

Question 25:

In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl?

Answer:

Suppose S represents a student chosen randomly studying in class XII and G represents a female student chosen randomly.We have, PG=4301000 PS/G=431000Now,PS/G=PSG PG=4310004301000=110                                                                         

Page No 31.35:

Question 26:

Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Answer:

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Consider the given events.
A = Even number appears on the card
B = A number, which is more than 3, appears on the card

Here,
A = {2, 4, 6, 8, 10}
B = {4, 5, 6, 7, 8, 9, 10}

Now, AB=4,6,8,10 Required probability=PA/B=nABnB=47

Page No 31.35:

Question 27:

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, then what is the constitutional probability that both are girls? Given that

(i) the youngest is a girl                                                 (b) at least one is a girl.                                                                                [CBSE 2014]

Answer:

Let G and B represents respectively a girl and a boy.

As, the possible outcomes of a family having two children is given by the set, S = {GG, GB, BG, BB}, where first letter represents the elder child.

So, n(S) = 4

Now,

i Let:A be the event that the family has two girl children andB be the event that the family has a girl as the youngest child.A=GG i.e. nA=1,B=GG,BG i.e. nB=2 andAB=GG i.e. nAB=1So, PA|B=nABnB=12ii Let:A be the event that the family has two girl children andB be the event that the family has at least one girl child.A=GG i.e. nA=1,B=GG,BG,GB i.e. nB=3 andAB=GG i.e. nAB=1So, PA|B=nABnB=13



Page No 31.53:

Question 1:

A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
(i) A = the first throw results in head, B = the last throw results in tail
(ii) A = the number of heads is odd, B = the number of tails is odd
(iii) A = the number of heads is two, B = the last throw results in head

Answer:

S=H H H  H H T H T H H T T T H H T H T T T H T T Ti PA=48=12PB=48=12Now, PAB=28=14 PAB=PAPBThus, A and B are independent events.ii PA=48=12PB=48=12Now, PAB=08=0 PAB  PAPBThus, A and B are not independent events.iii PA=38PB=48=12Now, PAB=28=14 PAB  PAPBThus, A and B are not independent events.

Page No 31.53:

Question 2:

Prove that in throwing a pair of dice, the occurrence of the number 4 on the first die is independent of the occurrence of 5 on the second die.

Answer:

Total number of events=36P4 on first die=PA=636=16P5 on second die=PB=636=16PAB=136PAB=PAPBThus, A and B are independent events.



Page No 31.54:

Question 3:

A card is drawn from a pack of 52 cards so the teach card is equally likely to be selected. In which of the following cases are the events A and B independent?
(i) A = The card drawn is a king or queen, B = the card drawn is a queen or jack
(ii) A = the card drawn is black, B = the card drawn is a king
(iii) B = the card drawn is a spade, B = the card drawn in an ace

Answer:

i Pking or queen=PA=852=213Pqueen or jack= PB=852=213PAB=Pqueen=452=113PAB  PA PBThus, A and B are not independent events.ii Pblack=PA=2652=12Pking= PB=452=113PAB=Pblack king=252=126PAB = PA PBThus, A and B are independent events.iii Pspade=PA=1352=14Pace= PB=452=113PAB=Pace of spade=152PAB =PA PBThus, A and B are independent events.

Page No 31.54:

Question 4:

A coin is tossed three times. Let the events A, B and C be defined as follows:
A = first toss is head, B = second toss is head, and C = exactly two heads are tossed in a row.
Check the independence of (i) A and B (ii) B and C and (iii) C and A

Answer:

S=H H H H H T H T H H T T T H H T H T T T H T T Ti PA=48=12PB=48=12PAB=28=14=PAPBThus, A and B are independent events.ii PC=28=14PB=48=12PBC=28=14 PBPCThus, B and C are not independent events.iii PC=14PA=12 PCA=18=PCPAThus, A and C are independent events.

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Question 5:

If A and B be two events such that P (A) = 1/4, P (B) = 1/3 and P (AB) = 1/2, show that A and B are independent events.

Answer:

PAB =PA+ PB-PABPAB =PA+ PB-PABPAB =14+13-12PAB =3+4-612PAB =112=14×13=PAPBThus, A and B are independent events.

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Question 6:

Given two independent events A and B such that P (A) = 0.3 and P (B) = 0.6. Find
(i) P (AB)
(ii) P (AB)
(iii) P (AB)
(iv) P AB
(v) P (AB)
(vi) P (A/B)
(vii) P (B/A)

Answer:

Given: A and B are independent events.PA=0.3PB=0.6i PAB =PA PB=0.3×0.6=0.18ii PAB¯ =PA PB¯=PA1-PB=0.3×1-0.6=0.3×0.4=0.12iii PA¯B =PA¯ PB=PB1-PA=0.6×0.7=0.42iv PA¯B¯ =PA¯ PB¯=1-PA1-PB=0.7×0.4=0.28v PAB =PA+ PB-PAB=0.3+0.6-0.18=0.9-0.18=0.72vi PAB=PABPB=0.180.6=0.3vii PBA=PABPA=0.180.3=0.6

Page No 31.54:

Question 7:

If P (not B) = 0.65, P (AB) = 0.85, and A and B are independent events, then find P (A).

Answer:

PB¯=0.651-PB=0.65PB=1-0.65=0.35Now,PAB=PA+PB-PABPAB=PA+PB-PA×PB0.85=PA+0.35-0.35×PA0.85-0.35=PA1-0.35PA=0.50.65=0.77

Page No 31.54:

Question 8:

If A and B are two independent events such that P (AB) = 2/15 and P (AB) = 1/6, then find P (B).

Answer:

Let:PA=x PB=yPA¯ B=215PA¯ ×P B=2151-xy=215        ...1PA B¯=16PA ×P B=16 1-yx=16        ...2Subtracting 2 from 1, we getx-y=130x=y+130Substituting the value of x in 2, we gety+1301-y=1630y2-29y+4=0y=16, 45

Page No 31.54:

Question 9:

A and B are two independent events. The probability that A and B occur is 1/6 and the probability that neither of them occurs is 1/3. Find the probability of occurrence of two events.

Answer:

PAB=PA PB        A and B are independent events16=PA PBPA=16PB          ...1PA¯B¯=1-PA1-PB13=1-PA1-PB13=1-16PB1-PB                 Using 1Let PB=x6x-16x1-x=136x-1-6x2+x=2x6x2-5x+1=02x-13x-1=0x=12 or x=13If PB=12, then PA=13If PB=13, then PA=12

Page No 31.54:

Question 10:

If A and B are two independent events such that P (AB) = 0.60 and P (A) = 0.2, find P (B).

Answer:


PAB =P A + PB-PABPAB =P A + PB-P A×PB        A and B are independent events0.6=0.2 +PB-0.2×PB0.6-0.2=PB1-0.2PB=0.6-0.21-0.2PB=0.40.8PB=12PB=0.5

Page No 31.54:

Question 11:

A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.

Answer:

S=  {1 1, 1 2, 1 3, 1 4, 1 5, 1 6,         2 1, 2 2, 2 3, 2 4, 2 5, 2 6,         3 1, 3 2, 3 3, 3 4, 3 5, 3 6,         4 1, 4 2, 4 3, 4 4, 4 5, 4 6,         5 1, 5 2, 5 3, 5 4, 5 5, 5 6,         6 1, 6 2, 6 3, 6 4, 6 5, 6 6}nS= 36E=Getting a number greater than 3 on each toss    =4 4, 4 5, 4 6, 5 4, 5 5, 5 6, 6 4, 6 5, 6 6nE=9PE=936       =14

Page No 31.54:

Question 12:

Given the probability that A can solve a problem is 2/3 and the probability that B can solve the same problem is 3/5. Find the probability that none of the two will be able to solve the problem.

Answer:

PA solving the problem=PA=23PB solving the problem=PB=35We need to find out if PA¯ B¯=PA¯ PB¯      A and B are independent events =1-PA1-PB=1-231-35=13×25=215

Page No 31.54:

Question 13:

An unbiased die is tossed twice. Find the probability of getting 4, 5, or 6 on the first toss and 1, 2, 3 or 4 on the second toss.

Answer:

PA=P4, 5 or 6 on first toss=1836=12PB=P1, 2, 3 or 4 on second toss=2436=23It is clear that A and B are independent events.PAB=PAPBPAB=12×23 PAB=13

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Question 14:

A bag contains 3 red and 2 black balls. One ball is drawn from it at random. Its colour is noted and then it is put back in the bag. A second draw is made and the same procedure is repeated. Find the probability of drawing (i) two red balls, (ii) two black balls, (iii) first red and second black ball.

Answer:

Given: Bag contains 3 red and 2 black balls.Let three red balls be R1, R2 and R3 and 2 black balls be B1 and B2.Sample space: {R1,R2, R1,R3, R1, B1, R1, B2, R1, B3R2, R3, R2, B1, R2, B2R3, R3iPdrawing two red balls =925iiPdrawing two black = 425iiiPfirst red and second black = 625

Page No 31.54:

Question 15:

Three cards are drawn with replacement from a well shuffled pack of cards. Find the probability that the cards drawn are king, queen and jack.

Answer:

Pking=452Pqueen=452Pjack=452These cards can be drawn in 3P3 ways.Pking, queen and jack=452×452×452×3P3=3!2197=62197

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Question 16:

An article manufactured by a company consists of two parts X and Y. In the process of manufacture of the part X, 9 out of 100 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part Y. Calculate the probability that the assembled product will not be defective.

Answer:

Let:A=Particle X is defectiveB=Particle Y is defective P(A)=9100     P(B)=5100Required probability=PA¯B¯                             =PA¯×PB¯                             =1-PA×1-PB                             =1-9100×1-5100                             =91100×95100                             =0.91×0.95                             =0.8645

Page No 31.54:

Question 17:

The probability that A hits a target is 1/3 and the probability that B hits it, is 2/5, What is the probability that the target will be hit, if each one of A and B shoots at the target?

Answer:

PA=PA hits target=13PB=PB hits target=25Now,PAB=Ptarget will be hit by either A or BPAB=PA+PB-PAB PAB=PA+PB-PAPB            A and B are independentPAB=13+25-13×25PAB=5+6-215PAB=915PAB=35

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Question 18:

An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane?

Answer:

P gun hits the plane=1-gun does not hit the planePA=1-PA¯Now,PA¯=1-0.41-0.31-0.21-0.1           =0.6×0.7×0.8×0.9           =0.3024 PA=1-0.3024           =0.6976

Page No 31.54:

Question 19:

The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that (i) at least one of the events will occur, and (ii) none of the events will occur.

Answer:

The odds against event A are 5 to 2.PA = 25+2=27The odds in favour of event B are 6 to 5.PB=66+5=611i Patleast one event occurs =PAB=PA+PB-PAB=PA+PB-PA×PB=27+611-27×611=22+4277-1277=22+42-1277=5277 PAB=5277ii Pnone of the event occurs=1-PAB=1-5277=2577

Page No 31.54:

Question 20:

A die is thrown thrice. Find the probability of getting an odd number at least once.

Answer:

Pgetting an odd number in one throw=12Here, getting an odd number in three throws refers to 3 independent events.PA=PB=PC=12PABC=PA+PB+PC-PAB+BC+CA+PABC=12+12+12-12×12+12×12+12×12+12×12×12=32-34+18=12-6+18=78

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Question 21:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red, (ii) first ball is black and second is red, (iii) one of them is black and other is red.

Answer:

Total balls =10 black + 8 red balls = 18 ballsPfirst red ball = 818Psecond red ball = 818Pfirst ball is black=1018Psecond ball is black=1018i Ptwo red balls= 818×818=1681ii Pfirst ball is black and second is red= 1018×818=2081iii Pone of them is black and other is red= Pfirst ball is red and second is black+Pfirst ball is black and second is red=818×1018+1018×818=2081+2081=4081

Page No 31.54:

Question 22:

An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting (i) 2 red balls, (ii) 2 blue balls, (iii) one red and one blue ball.

Answer:

Total balls=4 red balls +7 blue balls=11 ballsi P2 red balls= Pfirst ball is red×Psecond ball is red=411×411=16121ii P2 blue balls=Pfirst ball is blue×Psecond ball is blue=711×711=49121iii Pone red and one blue=Pfirst red and second blue+Pfirst blue and second red=411×711+711×411=28121+28121=56121Disclaimer: In the question, instead of black balls it should be blue balls.



Page No 31.55:

Question 23:

The probabilities of two students A and B coming to the school in time are 37and57 respectively. Assuming that the events, 'A coming in time' and 'B coming in time' are independent, find the probability of only one of them coming to the school in time. Write at least one advantage of coming to school in time.

Answer:

PA coming in time = 37PA not coming in time = 1-37=47PB coming in time = 57PB not coming in time = 1-57=27Ponly one of A and B coming in time = PA PB¯ + PA¯PB=37×27+47×57=649+2049=2649

Page No 31.55:

Question 24:

Two dice are thrown together and the total score is noted. The event E, F and G are "a total of 4", "a total of 9 or more", and "a total divisible by 5", respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.               [NCERT EXEMPLAR]

Answer:

We have,

S =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(S) = 36

E = "a total of 4" = {(1, 3), (2, 2), (3, 1)} i.e. n(E) = 3
F = "a total of 9 or more = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)} i.e. n(F) = 10
G = "a total divisible by 5" = {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)} i.e. n(G) = 7

Now,

PE=336=112,PF=1036=518 andPG=736Also,EF=ϕ, EG=ϕ and FG=4,6,5,5,6,4 i.e. nFG=3Since, PF×PG=518×736=35648336i.e. PF×PGPFGSo, no pair is independent.

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Question 25:

Let A and B be two independent events such that P(A) = p1 and P(B) = p2. Describe in words the events whose probabilities are:

(i) p1p2               (ii) (1 - p1)p2               (iii) 1 - (1 - p1)(1 - p2)               (iv) p1p2 - 2p1p2                       [NCERT EXEMPLAR]

Answer:

i As, p1p2=PA×PBAnd, A and B are independent events.i.e. PA×PB=PABSo, PAB=p1p2Hence, p1p2=PA and B occurii As, 1-p1p2=1-PA×PB=PA×PBAnd, A and B are independent events.i.e. PA×PB=PABSo, PAB=1-p1p2Hence, 1-p1p2=P A does not occur, but B occursiii As, 1-1-p11-p2=1-1-PA×1-PB=1-PA×PBAnd, A and B are independent events.i.e. PA×PB=PAB1-1-p11-p2=1-PAB=1-PAB=PABSo, PAB=1-1-p11-p2Hence, 1-1-p11-p2=PAt least one of A and B occursiv As, p1+p2-2p1p2=p1-p1p2+p2-p1p2=PA-PA×PB+PB-PA×PBAnd, A and B are independent events.i.e. PA×PB=PABp1+p2-2p1p2=PA-PAB+PB-PAB=Ponly A+Ponly BSo, Ponly A+Ponly B=p1+p2-2p1p2Hence, p1+p2-2p1p2=PExactly one of A and B occurs



Page No 31.68:

Question 1:

A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour.

Answer:

Given:Bag 1=3W+6B ballsBag 2=5B+4W ballsPballs of same colour are drawn=Pboth black+Pboth white=69×59+39×49=3081+1281=4281=1427

Page No 31.68:

Question 2:

A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.

Answer:

Given:Bag 1=3R+5B ballsBag 2=6R+4B ballsPone is red and one is black=Pred from bag 1 and black from bag 2+Pred from bag 2 and black from bag 1=38×410+58×610=1280+3080=4280=2140

Page No 31.68:

Question 3:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both the balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.

Answer:

Given:Box=10B+8R ballsi Pboth red balls=818×818=64324=1681ii Pfirst black and second red=818×1018=80 324=2081iii Pone is red and one is black=Pfirst red and second black +Pfirst red and second black=8'18×1018+1018×818=80324+80324=4081

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Question 4:

Two cards are drawn successively without replacement from a well-shuffled deck of 52 cards. Find the probability of exactly one ace.

Answer:

Pexactly one ace=Pfirst card is ace +Psecond card is ace=452×4851+4852×451=192+19252×51=38452×51=32221

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Question 5:

A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?

Answer:

PBoth narrating different incident=PB lies and A speaks the truth+PA lies and B speaks the truth=PAB¯+PA¯B=PAPB¯+PA¯PB=0.751-0.8 +1-0.750.8=0.75×0.2+0.25×0.8=0.15+0.2=0.35=35%

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Question 6:

Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is 1/3 and that of Monika's selection is 1/5. Find the probability that
(i) both of them will be selected
(ii) none of them will be selected
(iii) at least one of them will be selected
(iv) only one of them will be selected.

Answer:

PKamal gets selected= PA=13PMonica gets selected= PB=15i Pboth get selected =PA×PB=13×15=115ii Pnone of them get selected=PA¯×PB¯=1-PA1-PB=1-131-15=23×45=815iii Patleast one of them gets selected = PAB=PA+PB-PAB=PA+PB-PA×PB=13+15-13×15=13+15-115=715iv Pone of them gets selected=PA¯PB+PB¯PA=PB1-PA+PA1-PB=151-13+131-15=215+415=615=25



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Question 7:

A bag contains 3 white, 4 red and 5 black balls. Two balls are drawn one after the other, without replacement. What is the probability that one is white and the other is black?

Answer:

Given: Box= 3 W+4R+5B ballsPone white and one black=Pfirst white and second black+Pfirst black and second white=312×511+512×311=15132+15132=30132=522

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Question 8:

A bag contains 8 red and 6 green balls. Three balls are drawn one after another without replacement. Find the probability that at least two balls drawn are green.

Answer:

Patleast 2 balls are green=1-Pat most one ball is green=1-Pfirst green+Psecond green+Pthird green+Pno green=1-614×813×712+814×613×712+814×713×612+814×713×612=1-3362184+3362184+3362184+3362184=1-13442184=8402184=513

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Question 9:

Arun and Tarun appeared for an interview for two vacancies. The probability of Arun's selection is 1/4 and that to Tarun's rejection is 2/3. Find the probability that at least one of them will be selected.

Answer:

PArun gets selected=PA=14PTarun gets rejected=PB¯=23PTarun gets selected=1-23=13Patleast one of them is selected=PABPAB=PA+PB-PAB=PA+PB-PA×PB=14+13-14×13=3+4-112=12

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Question 10:

A and B toss a coin alternately till one of them gets a head and wins the game. If A starts the game, find the probability that B will win the game.

Answer:

PB winning the game=Phead at 2nd turn+Phead at 4th turn+ ...=12×12+12×12×12×12+ ...=122+124+126+128+ ...=141+122+124+126+ ... =1411-14                     For infinite GP: 1+a+a2+a3+ ... =11-a=14×43=13

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Question 11:

Two cards are drawn from a well shuffled pack of 52 cards, one after another without replacement. Find the probability that one of these is red card and the other a black card?

Answer:

Pone red and one black=Pfirst red and second black+Pfirst black and second red=2652×2651+2652×2651                   Without replacement=1351+1351=2651

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Question 12:

Tickets are numbered from 1 to 10. Two tickets are drawn one after the other at random. Find the probability that the number on one of the tickets is a multiple of 5 and on the other a multiple of 4.

Answer:

We know that 5 and 10 are multiples of 5, while 4 and 8 are multiples of 4.Pmultiple of 5=210=15Pmultiple of 4=210=15Pmultiple of 5 and multiple of 4=Pmultiple of 5 on first card and multiple of 4 on second card                                               +Pmultiple of 4 on first card and multiple of 5 on second card=210×29+210×29              Without replacement=490+490=890=445

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Question 13:

In a family, the husband tells a lie in 30% cases and the wife in 35% cases. Find the probability that both contradict each other on the same fact.

Answer:

It is given that the husband lies in 30% of the cases, while the wife lies in 35% cases

Pboth will contradict each other on the same fact=Phusband lies but wife tells the truth+Pwife lies but husband tells the truth=0.3×1-0.35+1-0.3×0.35=0.3×0.65+0.7×0.35=0.195+0.245=0.44=44%

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Question 14:

A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is 1/7 and that of wife's selection is 1/5. What is the probability that
(i) both of them will be selected?
(ii) only one of them will be selected?
(iii) none of them will be selected?

Answer:

Phusband will be selected=PA=17Pwife will be selected=PB=15i Pboth will be selected=PAB=PA×PB=17×15=135ii Ponly one of them will be selected=PAPB¯+PA¯PB=171-15+151-17=435+635=1035=27iii Pnone of them will be selected=PAB=PA¯×PB¯=1-171-15=2435

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Question 15:

A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.

Answer:

Given: Bag=7W+5B+4R ballsPatleast 3 balls are black=Pexactly 3 black+Pall 4 black=1116×515×414×313×4+516×415×314×213=1114×13+12×14×13=22+1364=23364

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Question 16:

A, B, and C are independent witness of an event which is known to have occurred. A speaks the truth three times out of four, B four times out of five and C five times out of six. What is the probability that the occurrence will be reported truthfully by majority of three witnesses?

Answer:

PA speaks truth=34PB speaks truth=45PC speaks truth=56Pmajority speaks truth=Ptwo speak truth+Pall speak truth=PA×PB1-PC+PA×PC1-PB+PC×PB1-PA+PA×PB×PC=34×451-56+34×561-45+45×561-34+34×45×56=12120+15120+20120+60120=107120

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Question 17:

A bag contains 4 white balls and 2 black balls. Another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that
(i) both are white
(ii) both are black
(iii) one is white and one is black

Answer:

Given:Bag 1=4W+2B ballsBag 2=3W+5B ballsi Pboth are white=46×38=14ii Pboth are black=26×58=524iii Pone is white and one is black=Pwhite from bag 1 and black from bag 2+Pwhite from bag 2 and black from bag 1=46×58+38×26=2048+648=2648=1324

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Question 18:

A bag contains 4 white, 7 black and 5 red balls. 4 balls are drawn with replacement. What is the probability that at least two are white?

Answer:

Given: Bag=4W+5R+7B ballsPatleast 2 white balls=1-Pmaximum 1 white ball=1-Pno white+Pexactly one white=1-1216×1216×1216×1216+416×1216×1216×1216×4=1-81256+108256=1-189256=256-189256=67256

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Question 19:

Three cards are drawn with replacement from a well shuffled pack of 52 cards. Find the probability that the cards are a king, a queen and a jack.

Answer:

Pking=PA=452Pqueen=PB=452Pjack=PC=452Pking, queen and jack=3!×PA×PB×PC =3×2×452×452×452=62197

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Question 20:

A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, find the probability that the (i) balls are of different colours (ii) balls are of the same colour.

Answer:

Given:Bag A=4R+5B ballsBag B=3R+7B ballsi Pballs of different colours=Pred from bag A and black from bag B+Pred from bag B and black from bag A=49×710+310×59=2890+1590=4390ii Pballs of same colour=Pboth red +Pboth black=49×310+710×59=1290+3590=4790

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Question 21:

A can hit a target 3 times in 6 shots, B : 2 times in 6 shots and C : 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?

Answer:

PA hits the target=36PB hits the target=26PC hits the target=44=1Patleast 2 shots hit\=Pexactly 2 shots hit+Pall 3 shots hit=361-26+261-36+36×26×1               Here, the probability of C hitting the target is 1. So, it will  always hit.When exactly 2 shots are hit, then either A hits or B hits.=36×46+26×36+636=12+6+636=2436=23

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Question 22:

The probability of student A passing an examination is 2/9 and of student B passing is 5/9. Assuming the two events : 'A passes', 'B passes' as independent, find the probability of : (i) only A passing the examination (ii) only one of them passing the examination.

Answer:

PA passing examination=29PB passing examination=59i Ponly A passing examination=PA passes PB fails=291-59=29×49=881ii Ponly one of them passing examination=PA passes and B fails+PB passes and A fails=29×1-59+59×1-29=881+3581=4381

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Question 23:

There are three urns A, B, and C. Urn A contains 4 red balls and 3 black balls. urn B contains 5 red balls and 4 black balls. Urn C contains 4 red and 4 black balls. One ball is drawn from each of these urns. What is the probability that 3 balls drawn consists of 2 red balls and a black ball?

Answer:

Given:Urn A 4R+3BUrn B 5R+4BUrn C 4R+4BPtwo red and one black=Pblack from urn A+Pblack from urn B+Pblack from urn C=37×59×48+47×49×48+47×59×48=542×16126×20126=15+16+20126=51126=1742

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Question 24:

X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets
(i) Grade A in all subjects
(ii) Grade A in no subject
(iii) Grade A in two subjects.

Answer:

PA grade in Maths=PA=0.2PA grade in Physics=PB=0.3PA grade in Chemistry=PC=0.5i Pgrade A in all subjects=PA×PB×PC=0.2×0.3×0.5=0.03ii Pgrade A in no subject=PA¯×PB¯×PC¯=1-0.2×1-0.3×1-0.5=0.8×0.7×0.5=0.28iii Pgrade A in two subjects=Pnot grade A in Maths+Pnot grade A in Physics+Pnot grade A in Chemistry=PA×PB×PC+PA×PB×PC+PA×PB×PC=1-0.2×0.3×0.5+0.2×1-0.3×0.5+0.2×0.3×1-0.5=0.8×0.3×0.5+0.2×0.7×0.5+0.2×0.3×0.5=0.12+0.07+0.03=0.22



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Question 25:

A and B take turns in throwing two dice, the first to throw 9 being awarded the prize. Show that their chance of winning are in the ratio 9:8.

Answer:

Total number of events =36Pgetting 9=436=19PA winning=Pgetting 9 in first throw+Pgetting 9 in third throw+ ...=19+1-191-19×19+ ...=191+6481+64812+ ... =1911-6481                                     1+a+a2+a3+ ... =11-a=19×8117=917PB winning=Pgetting 9 in second throw+Pgetting 9 in fourth throw+ ...=1-1919+1-191-191-19×19+ ...=8811+6481+64812+ ... =88111-6481                                   1+a+a2+a3+ ... =11-a=881×8117=817 Winning ratio of A to B=917817=98

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Question 26:

A, B and C in order toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely?

Answer:

PA winning=Phead in first toss+Phead in fourth toss+ ...=12+12×12×12×12+ ...=121+123+126+ ... =1211-18                                   1+a+a2+a3+ ... =1 1-a=12×87=47PB winning=Phead in second toss+Phead in fifth toss+ ...=12×12+12×12×12×12×12+ ...=141+123+126+ ... =1411-18                                    1+a+a2+a3+ ... =11-a=14×87=27PC winning=Phead in third toss+Phead in sixth toss+ ...=12×12×12+12×12×12×12×12×12+ ...=181+123+126+ ... =1811-18                                1+a+a2+a3+ ... =11-a=18×87=17

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Question 27:

Three persons A, B, C throw a die in succession till one gets a 'six' and wins the game. Find their respective probabilities of winning.

Answer:

Psix=16Pno six=56PA winning=P6 in first throw+P6 in fourth throw+ ...=16+56×56×56×16+ ...=161+563+566+ ... =1611-125216                          1+a+a2+a3+ ... =11-a=16×21691=3691PB winning=P6 in second throw+P6 in fifth throw+ ...=56×16+56×56×56×56×16+ ...=5361+563+566+ ...=53611-125216                         1+a+a2+a3+ ... =11-a=536×21691=3091PC winning=P6 in third throw+P6 in sixth throw+ ...=56×56×16+56×56×56×56×56×12+ ...=252161+563+566+ ... =2521611-125216                         1+a+a2+a3+ ... =11-a=25216×21691=2591

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Question 28:

A and B take turns in throwing two dice, the first to throw 10 being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio 12 : 11.

Answer:

There are only three possible cases, wherein the sum of the numbers obtained after throwing 2 dice is 10, i.e. 4, 6 5, 5 6, 4. Psum of the numbers is 10=336=112Psum of the numbers is not 10=1-112=1112Pany numner other than six=56PA winning=P10 in first throw+P10 in third throw+ ...=112+1112×1112×112+ ...=1121+11122+11124+ ... =11211-121144                         1+a+a2+a3+ ... =11-a=112×14423=1223PB winning=1-PA winning=1123Now,PA winningPB winning=12231123=1211Hence proved.

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Question 29:

There are 3 red and 5 black balls in bag 'A'; and 2 red and 3 black balls in bag 'B'. One ball is drawn from bag 'A' and two from bag 'B'. Find the probability that out of the 3 balls drawn one is red and 2 are black.

Answer:

It is given that bag A contains 3 red and 5 black balls 3R, 5B and bag B contains 2 red and 3 black balls 2R, 3B.Now,Pone red and 2 black= Pone red from bag A and two black from bag B+Pblack ball from bag A and remaining balls from bag B=38×35×24+58×25×34×2=980+3080=3980Note: 2 is multiplied by second term because there are two ways to select red and black balls from bag B. While the first way is to pick black ball first, followed by red, the second way is to pick red ball first, followed by black.

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Question 30:

Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is 17 and that of John's selection is 15. What is the probability that
(i) both of them will be selected?
(ii) only one of them will be selected?
(iii) none of them will be selected?

Answer:

PFatima gets selected=PA=17PJohn gets selected=PB=15i Pboth of them get selected=PAB=PA×PB=17×15=135ii Ponly one of them gets selected=PA×PB¯+PA¯×PB=171-15+1-1715=17×45+67×15=435+635=1035=27iii Pnone of them get selected=PB¯×PA¯=1-15×1-17=45×67=2435

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Question 31:

A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its colour is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be
(i) blue followed by red.
(ii) blue and red in any order.
(iii) of the same colour.

Answer:

It is given that the bag contains 3 blue and 5 red marbles.i Pblue followed by red=38×58=1564ii Pred and blue in any order=Pblue followed by red+Pred followed by blue=38×58+58×38=1564+1564=3064=1532iii Psame colour=Pboth red+Pboth blue=58×58+38×38=2564+964=3464=1732

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Question 32:

An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting
(i) 2 red balls
(ii) 2 blue balls
(iii) One red and one blue ball.

Answer:

It is given that the urn contains 7 red and 4 black balls.i P2 red balls=711×711=49121ii P2 blue balls=411×411=16121iii Pone red ball and one blue ball=Pblue ball followed by red ball+Pred ball followed by blue ball=411×711+711×411=28121+28121=56121

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Question 33:

A card is drawn from a well-shuffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
(i) What is the probability that both the cards are of the same suit?
(ii) What is the probability that the first card is an ace and the second card is a red queen?

Answer:

i Pboth the cards are of same suit=Pboth the cards are of diamond+Pboth the cards are of spade+Pboth the cards are of club+Pboth the cards are of heart=1352×1352+1352×1352+1352×1352+1352×1352=116+116+116+116=416=14ii Pfirst ace and second red queen=Pace card×Pred queen=452×252=1338

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Question 34:

Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that: (i) you both enter the same section? (ii) you both enter the different sections?

Answer:

i Pboth enter same section=Pboth enter section A+Pboth enter section B=40100×40100+60100×60100=425+925=1325ii Pboth enter different sections=1-Pboth enter same section=1-1325=1225

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Question 35:

In a hockey match, both teams A and B scored same number of goals upto the end of the game, so to decide the winner, the refree asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the refree was fair or not.

Answer:

Pa six=16Pnot a six=1-16=56PA wins=P6 in first throw+P6 in third throw+ ...=16+56×56×16+ ...=161+562+564+ ...=1611-2536 ... 1+a+a2+a3... =11-a=16×3611=611PB wins=P6 in second throw+P6 in fourth throw+ ...=56×16+56×56×56×16+ ...=5361+562+564+ ...=53611-2536 ... 1+a+a2+a3... =11-a=536×3611=511

It can be seen that the probability that team A wins is not equal to the probability that team B wins.
Thus, the decision of the referee was not fair.

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Question 36:

A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins.

Answer:

Total of 7 on the dice can be obtained in the following ways:

(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)

Probability of getting a total of 7 = 636=16

Probability of not getting a total of 7 = 1-16=56

Total of 10 on the dice can be obtained in the following ways:

(4, 6), (6, 4), (5, 5)

Probability of getting a total of 10 = 336=112

Probability of not getting a total of 10 = 1-112=1112

Let E and F be the two events, defined as follows:

E = Getting a total of 7 in a single throw of a dice

F = Getting a total of 10 in a single throw of a dice

P(E) = 16, PE¯=56, P(F) = 112, PF¯=1112

A wins if he gets a total of 7 in 1st, 3rd or 5th ... throws.

Probability of A getting a total of 7 in the 1st throw = 16

A will get the 3rd throw if he fails in the 1st throw and B fails in the 2nd throw.

Probability of A getting a total of 7 in the 3rd throw = PE¯ PF¯ PE=56×1112×16

Similarly, probability of getting a total of 7 in the 5th throw = PE¯ PF¯ PE¯ PF¯ PE=56×1112×56×1112×16 and so on
Probability of winning of A = 16+ 56×1112×16+ 56×1112×56×1112×16+...  =161-56×1112=1217
∴ Probability of winning of B = 1 − Probability of winning of A =  1-1217=517



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Question 1:

A bag A contains 5 white and 6 black balls. Another bag B contains 4 white and 3 black balls. A ball is transferred from bag A to the bag B and then a ball is taken out of the second bag. Find the probability of this ball being black.

Answer:

A black ball can be drawn in two mutually exclusive ways:

(I) By transferring a white ball from bag A to bag B, then drawing a black ball
(II) By transferring a black ball from bag A to bag B, then drawing a black ball

Let E1, E2 and A be the events as defined below:

E1 = A white ball is transferred from bag A to bag B
E2 = A black ball is transferred from bag A to bag B
A = A black ball is drawn

 PE1=511      PE2=611Now, PA/E1=38PA/E2=48Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =511×38+611×48                                                   =1588+2488                                                   =3988

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Question 2:

A purse contains 2 silver and 4 copper coins. A second purse contains 4 silver and 3 copper coins. If a coin is pulled at random from one of the two purses, what is the probability that it is a silver coin?

Answer:

A silver coin can be drawn in two mutually exclusive ways:

(I) Selecting purse I and then drawing a silver coin from it
(II) Selecting purse II and then drawing a silver coin from it

Let E1, E2 and A be the events as defined below:

E1 = Selecting purse I
E2 = Selecting purse II
A = Drawing a silver coin

It is given that one of the purses is selected randomly.

 PE1=12      PE2=12Now, PA/E1=26=13PA/E2=47Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =12×13+12×47                                                   =16+27                                                   =7+1242=1942

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Question 3:

One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.

Answer:

A yellow ball can be drawn in two mutually exclusive ways:

(I) By transferring a red ball from first to second bag, then drawing a yellow ball
(II) By transferring a yellow ball from first to second bag, then drawing a yellow ball

Let E1, E2 and A be the events as defined below:

E1 = A red ball is transferred from first to second bag
E2 = A yellow ball is transferred from first to second bag
A =  A yellow ball is drawn

 PE1=59      PE2=49Now, PA/E1=610PA/E2=710Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =59×610+49×710                                                   =3090+2890                                                   =5890=2945

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Question 4:

A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.

Answer:

A white ball can be drawn in two mutually exclusive ways:

(I) Selecting bag I and then drawing a white ball from it
(II) Selecting bag II and then drawing a white ball from it

Let E1, E2 and A be the events as defined below:
E1 = Selecting bag I
E2 = Selecting bag II
A = Drawing a white ball

It is given that one of the bags is selected randomly.

 PE1=12      PE2=12Now,PA/E1=35PA/E2=26=13Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =12×35+12×13                                                   =310+16                                                   =9+530=1430=715

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Question 5:

The contents of three bags I, II and III are as follows:
Bag I : 1 white, 2 black and 3 red balls,
Bag II : 2 white, 1 black and 1 red ball;
Bag III : 4 white, 5 black and 3 red balls.
A bag is chosen at random and two balls are drawn. What is the probability that the balls are white and red?

Answer:

A white ball and a red ball can be drawn in three mutually exclusive ways:

(I) Selecting bag I and then drawing a white and a red ball from it
(II) Selecting bag II and then drawing a white and a red ball from it
(II) Selecting bag III and then drawing a white and a red ball from it

Let E1, E2 and A be the events as defined below:
E1 = Selecting bag I
E2 = Selecting bag II
E3 = Selecting bag II
A = Drawing a white and a red ball

It is given that one of the bags is selected randomly.

 PE1=13     PE2=13     PE3=13Now, PA/E1=1C1×3C16C2=315PA/E2=2C1×1C14C2=26PA/E3=4C1×3C112C2=1266Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2+ PE3PA/E3                                                   =13×315+13×26+13×1266                                                   =115+19+233                                                   =33+55+30495=118495

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Question 6:

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered 2, 3, 4, ..., 12 is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?

Answer:

Let E1, E2 and A be the events as defined below:
E1 = The coin shows a head
E2 = The coin shows a head
A = The noted number is 7 or 8

 PE1=12      PE2=12Now, PA/E1=1136PA/E2=211Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =12×1136+12×211                                                   =1172+111                                                   =121+72792=193792

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Question 7:

A factory has two machines A and B. Past records show that the machine A produced 60% of the items of output and machine B produced 40% of the items. Further 2% of the items produced by machine A were defective and 1% produced by machine B were defective. If an item is drawn at random, what is the probability that it is defective?

Answer:

Let A, E1 and E2 denote the events that the item is defective, machine A is selected and machine B is selected, respectively.
 
 PE1=60 100      PE2=40100Now,PA/E1=2100PA/E2=1100Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =60100×2100+40100×1100                                                   =12010000+4010000                                                   =120+4010000=16010000=0.016

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Question 8:

The bag A contains 8 white and 7 black balls while the bag B contains 5 white and 4 black balls. One ball is randomly picked up from the bag A and mixed up with the balls in bag B. Then a ball is randomly drawn out from it. Find the probability that ball drawn is white.

Answer:

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from bag A to bag B, then drawing a white ball
(II) By transferring a white ball from bag A to bag B, then drawing a white ball

Let E1, E2 and A be events as defined below:
E1 = A black ball is transferred from bag A to bag B
E2 = A white ball is transferred from bag A to bag B
A = A white ball is drawn

 PE1=715     PE2=815Now,PA/E1=510=12PA/E2=610=35Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =715×12+815×35                                                   =730+825                                                   =35+48150=83150



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Question 9:

A bag contains 4 white and 5 black balls and another bag contains 3 white and 4 black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from the latter. Find the probability that the ball drawn is white.

Answer:

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from first to second bag, then drawing a white ball
(II) By transferring a white ball from first to second bag, then drawing a white ball

Let E1, E2 and A be the events as defined below:
E1 = A black ball is transferred from first to second bag
E2 = A white ball is transferred from first to second bag
A = A white ball is drawn

 PE1=59      PE2=49Now,PA/E1=38PA/E2=48=12Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =59×38+49×12                                                   =1572+29                                                   =15+1672=3172

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Question 10:

One bag contains 4 white and 5 black balls. Another bag contains 6 white and 7 black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is white.

Answer:

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from first to second bag, then drawing a white ball
(II) By transferring a white ball from first to second bag, then drawing a white ball

Let E1, E2 and A be the events as defined below:
E1 = A black ball is transferred from first to second bag
E2 = A white ball is transferred from first to second bag
A = A white ball is drawn

 PE1=59      PE2=49Now, PA/E1=614PA/E2=714Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =59×614+49×714                                                   =30126+28126                                                   =58126=2963

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Question 11:

An urn contains 10 white and 3 black balls. Another urn contains 3 white and 5 black balls. Two are drawn from first urn and put into the second urn and then a ball is drawn from the latter. Find the probability that its is a white ball.

Answer:

A white ball can be drawn in three mutually exclusive ways:

(I) By transferring two black balls from first to second urn, then drawing a white ball
(II) By transferring two white balls from first to second urn, then drawing a white ball
(III) By transferring a white and a black ball from first to second urn, then drawing a white ball

Let E1, E2, E3 and A be the events as defined below:
E1 = Two black balls are transferred from first to second bag
E2 = Two white balls are transferred from first to second bag
E2 = A white and a black ball is transferred from first to second bag
A = A white ball is drawn

 PE1=3C213C2=378 PE2=10C213C2=4578 PE3=10C1×3C113C2=3078Now, PA/E1=310PA/E2=510PA/E3=410Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2+PE3PA/E3                                                   =378×310+4578×510+3078×410                                                   =9780+225780+120780                                                   =354780=59130

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Question 12:

A bag contains 6 red and 8 black balls and another bag contains 8 red and 6 black balls. A ball is drawn from the first bag and without noticing its colour is put in the second bag. A ball is drawn from the second bag. Find the probability that the ball drawn is red in colour.

Answer:

A red ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from first to second bag, then drawing a red ball
(II) By transferring a red ball from first to second bag, then drawing a red ball

Let E1, E2 and A be the events as defined below:
E1 = A black ball is transferred from first to second bag
E2 = A red ball is transferred from first to second bag
A = A red ball is drawn

 PE1=814      PE2=614Now,PA/E1=815PA/E2=915Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2                                                   =814×815+614×915                                                   =64210+54210                                                   =118210=59105

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Question 13:

Three machines E1, E2, E3 in a certain factory produce 50%, 25% and 25%, respectively, of the total daily output of electric bulbs. It is known that 4% of the tubes produced one each of the machines Eand E2 are defective, and that 5% of those produced on E3 are defective. If one tube is picked up at random from a day's production, then calculate the probability that it is defective.
                                                                                                                                                                      [NCERT EXEMPLAR, CBSE 2015]

Answer:

Let A be the event that the tube picked is defective.

We have,PE1=50%=50100=12, PE2=25%=25100=14, PE3=25%=25100=14,PA|E1=4%=4100=125, PA|E2=4%=4100=125 and PA|E3=5%=5100=120Now,PA=PE1×PA|E1+PE2×PA|E2+PE3×PA|E3=12×125+14×125+14×120=150+1100+180=8+4+5400=17400

So, the probability that the picked tube is defective is 17400.



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Question 1:

The contents of urns I, II, III are as follows:
Urn I : 1 white, 2 black and 3 red balls
Urn II : 2 white, 1 black and 1 red balls
Urn III : 4 white, 5 black and 3 red balls.
One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from Urns I, II, III?

Answer:

Let E1, E2 and E3 denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the two balls drawn are white and red.

 PE1=13     PE2=13      PE3=13Now, PA/E1=1C1×3C16C2=315=15PA/E2=2C1×1C14C2=26=13PA/E3=4C1×3C112C2=1266=211Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE3PA/E3                                                   =13×1513×15+13×13+13×211                                                   =1515+13+211=1533+55+30165=33118Required probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2++ PE3PA/E3                                                   =13×1313×15+13×13+13×211                                                   =1315+13+211=1333+55+30165=55118Required probability = PE3/A=PE3PA/E3PE1PA/E1+ PE2PA/E2++ PE3PA/E3                                                   =13×21113×15+13×13+13×211                                                   =21115+13+211=21133+55+30165=30118                                                   

 

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Question 2:

A bag A contains 2 white and 3 red balls and a bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag B.

Answer:

Let A, E1 and E2 denote the events that the ball is red, bag A is chosen and bag B is chosen, respectively.

 PE1=12      PE2=12Now, PA/E1=35PA/E2=59Using Bayes' theorem, we getRequired probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2                                                   =12×5912×35+12×59                                                   =2552                                                   



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Question 3:

Three urns contains 2 white and 3 black balls; 3 white and 2 black balls and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random and it was found to be white. Find the probability that it was drawn from the first urn.

Answer:

Let E1, E2 and E3 denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the ball drawn is white.

 PE1=13     PE2=13     PE3=13Now, PA/E1=25PA/E2=35PA/E3=45Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE3PA/E3                                                   =13×2513×25+13×35+13×45                                                   =22+3+4=29                                                                     

 

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Question 4:

The contents of three urns are as follows:
Urn 1 : 7 white, 3 black balls, Urn 2 : 4 white, 6 black balls, and Urn 3 : 2 white, 8 black balls. One of these urns is chosen at random with probabilities 0.20, 0.60 and 0.20 respectively. From the chosen urn two balls are drawn at random without replacement. If both these balls are white, what is the probability that these came from urn 3?

Answer:

Let E1, E2 and E3 denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the two balls drawn are white.

 PE1=20100     PE2=60100      PE3=20100Now,PA/E1=7C210C2=2145PA/E2=4C210C2=645PA/E3=2C210C2=145Using Bayes' theorem, we getRequired probability = PE3/A=PE3PA/E3PE1PA/E1+ PE2PA/E2++ PE3PA/E3                                                   =20100×14520100×2145+60100×645+20100×145                                                   =121+18+1=140                                                   

 

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Question 5:

Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one 'tail', then what is the probability that she threw 3, 4, 5 or 6 with the die?                                                                                                                                                                                    [CBSE 2015]

Answer:

Let E1 be the event that the outcome on the die is 1 or 2 and E2 be the event that the outcome on the die is 3, 4, 5 or 6. Then,

PE1=26=13 and PE2=46=23

Let A be the event of getting exactly one 'tail'.

P(A|E1) = Probability of getting exactly one tail by tossing the coin three times if she gets 1 or 2 = 38

P(A|E2) = Probability of getting exactly one tail in a single throw of a coin if she gets 3, 4, 5 or 5 = 12

As, the probability that the girl threw 3, 4, 5 or 6 with the die, if she obtained exactly one tail, is given by P(E2|A).

So, by using Baye's theorem, we get

PE2|A=PE2×PA|E2PE1×PA|E1+PE2×PA|E2=23×1213×38+23×12=2618+13=261124=24×211×6=811

So, the probability that she threw 3, 4, 5 or 6 with the die if she obtained exactly one tail is 811.

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Question 6:

Two groups are competing for the positions of the Board of Directors of a Corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Answer:

Let E1 and E2 denote the events that the first group and the second group win the competition, respectively. Let A be the event of introducing a new product.

P(E1) = Probability that the first group wins the competition = 0.6

P(E2) = Probability that the second group wins the competition = 0.4

P(A/E1) = Probability of introducing a new product if the first group wins = 0.7

P(A/E2) = Probability of introducing a new product if the second group wins = 0.3

The probability that the new product is introduced by the second group is given by P(E2/A).

Using Bayes’ theorem, we get
Required probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2                                                   =0.4×0.30.6×0.7+0.4×0.3                                                   =0.120.54=29                                                   

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Question 7:

Suppose 5 men out of 100 and 25 women out of 1000 are good orators. An orator is chosen at random. Find the probability that a male person is selected. Assume that there are equal number of men and women.

Answer:

Let A, E1 and E2 denote the events that the person is a good orator, is a man and is a woman, respectively.

 PE1=12      PE2=12Now,PA/E1=5100PA/E2=251000Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =12×510012×5100+12×251000                                                   =11+12=23                                                   

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Question 8:

A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from
(i) LONDON (ii) CLIFTON?

Answer:

Let A, E1 and E2 denote the events that the two consecutive letters are visible, the letter has come from LONDON and the letter has come from CLIFTON, respectively.

 PE1=12      PE2=12Now,PA/E1=25PA/E2=16Using Bayes' theorem, we geti Required probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =12×2512×25+12×16                                                   =2525+16=251730=1217ii Required probability = PE2/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =12×1612×25+12×16                                                   =1625+16=161730=517                                                   

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Question 9:

In a class, 5% of the boys and 10% of the girls have an IQ of more than 150. In this class, 60% of the students are boys. If a student is selected at random and found to have an IQ of more than 150, find the probability that the student is a boy.

Answer:

Let A, E1 and E2 denote the events that the IQ is more than 150, the selected student is a boy and the selected student is a girl, respectively.

 PE1=60100      PE2=40100Now, PA/E1=5100PA/E2=10100Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =60100×510060100×5100+40100×10100                                                   =300300+400=300700=37                                                   

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Question 10:

A factory has three machines X, Y and Z producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% and Z produces 2% defective bolts. At the end of a day, a bolt is drawn at random and is found to be defective. What is the probability that this defective bolt has been produced by machine X?

Answer:

Let E1, E2 and E3 denote the events that machine X produces bolts, machine Y produces bolts and machine Z produces bolts, respectively.

Let A be the event that the bolt is defective.

Total number of bolts = 1000 + 2000 + 3000 = 6000

P(E1) = 10006000=16

P(E2) =20006000=13

P(E3) = 30006000=12

The probability that the defective bolt is produced by machine X is given by P (E1/A).

Now, PA/E1=1%=1100PA/E2=1.5%=151000PA/E3=2%=2100Using Bayes' theorem, we get Required probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE2PA/E2                                                   =16×110016×1100+13×151000+12×2100                                                   =1616+12+1=161+3+66=110

 

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Question 11:

An insurance company insured 3000 scooters, 4000 cars and 5000 trucks. The probabilities of the accident involving a scooter, a car and a truck are 0.02, 0.03 and 0.04 respectively. One of the insured vehicles meet with an accident. Find the probability that it is a (i) scooter (ii) car (iii) truck.

Answer:

Let E1, E2 and E3 denote the events that the vehicle is a scooter, a car and a truck, respectively.

Let A be the event that the vehicle meets with an accident.

It is given that there are 3000 scooters, 4000 cars and 5000 trucks.

Total number of vehicles = 3000 + 4000 + 5000 = 12000

P(E1) = 300012000=14

P(E2) =400012000=13

P(E3) = 500012000=512

The probability that the vehicle, which meets with an accident, is a scooter is given by P (E1/A).

Now, PA/E1=0.02=2100PA/E2=0.03=3100PA/E3=0.04=4100Using Bayes' theorem, we geti Required probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE2PA/E2                                                   =14×210014×2100+13×3100+512×4100                                                   =1212+1+53=123+6+106=319ii Required probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2++ PE2PA/E2                                                   =13×310014×2100+13×3100+512×4100                                                   =112+1+53=13+6+106=619iii Required probability = PE2/A=PE3PA/E3PE1PA/E1+ PE2PA/E2++ PE2PA/E2                                                   =512×410014×2100+13×3100+512×4100                                                   =5312+1+53=533+6+106=1019                                                   

 

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Question 12:

Suppose we have four boxes A, B, C, D containing coloured marbles as given below:
Figure

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B? box C?

Answer:

Let R be the event of drawing the red marble.

Let EA, EB and EC denote the events of selecting box A, box B and box C, respectively.

Total number of marbles = 40

Number of red marbles = 15

   PR=1540=38

Probability of drawing a red marble from box A is given by P(EA/R).

  PEA/R=PEARPR=14038=115

Probability of drawing a red marble from box B is given by P(EB/R).

 PEB/R=PEBRPR=64038=25

Probability of drawing a red marble from box C is given by P(EC/R).

  PEC/R=PECRPR=84038=815

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Question 13:

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced. What is the probability that it was produced by A?

Answer:

Let E1, E2 and E3 be the time taken by machine operators A, B, and C, respectively.

Let X be the event of producing defective items.

 PE1=50%=12      PE2=30%=310     PE3=20%=15Now, PA/E1=1%=1100PA/E2=5%=5100PA/E2=7%=7100Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE2PA/E2                                                   =12×110012×1100+310×5100+15×7100                                                   =534                                                   

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Question 14:

An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on machine A, 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective and 3% of these produced on C are defective. All the items stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?                                                                                                  [NCERT EXEMPLAR]

Answer:

Let E be the event of getting a defective item.We have,PA=50%=50100=12, PB=30%=30100=310 and PC=20%=20100=15,PE|A=2%=2100=150, PE|B=2%=2100=150 and PE|C=3%=3100Now,Pthe defective item drawn was manufactured on machine A=PA×PE|APA×PE|A+PB×PE|B+PC×PE|C=12×15012×150+310×150+15×3100=11001100+3500+3500=11005+3+3500=110011500=500100×11=511

So, the probability that the defective item drawn was manufactured on machine A is 511.

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Question 25:

An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcycle is 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.

Answer:

Let A, E1 and E2 denote the events that the vehicle meets the accident, is a scooter and is a motorcycle, respectively.

 PE1=20005000=0.4      PE2=30005000=0.6Now,PA/E1=0.01PA/E2=0.02Using Bayes' theorem, we getRequired probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2                                                   =0.6×0.020.6×0.02+0.4×0.01                                                   =0.0120.012+0.004=0.0120.016=34                                                   



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Question :

A speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?

Answer:

Let A, E1 and E2 denote the events that the man reports the appearance of 5 on throwing a die, 5 appears and 5 does not appear, respectively.

 PE1=16     PE2=56Now, PA/E1=810PA/E2=210Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =16×81016×810+56×210                                                   =88+10=818=49                                                   

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Question 15:

There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tail 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?                                                                                                               [CBSE 2014]

Answer:

Let:A be the event of choosing two-headed coin,B be the event of choosing a biased coin that comes up head 75% of the times,C be the event of choosing a biased coin that comes up tail 40% of the times andE be the event of getting a head.Now,PA=PB=PC=13 andPE|A=1, PE|B=75%=75100=34 and PE|C=60%=60100=35So, using Bayes' theorem, we getPthe head shown was of two-headed coin=PA|E=PA×PE|APA×PE|A+PB×PE|B+PC×PE|C=13×113×1+13×34+13×35=1313+14+15=1320+15+1260=134760=603×47=2047

So, the probability that the head shown was of a two-headed coin is 2047.

Disclaimer: The answer given in the book is incorrect. The same has been corrected here.

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Question 16:

In a factory, machine A produces 30% of the total output, machine B produces 25% and the machine C produces the remaining output. If defective items produced by machines A, B and C are 1%, 1.2%, 2% respectively. Three machines working together produce 10000 items in a day. An item is drawn at random from a day's output and found to be defective. Find the probability that it was produced by machine B?

Answer:

Let A, E1, E2 and E3 denote the events that the item is defective, machine A is chosen, machine B is chosen and machine C is chosen, respectively.

 PE1=30100     PE2=25100      PE3=45100Now, PA/E1=1100PA/E2=1.2100PA/E3=2100Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2+PE3PA/E3                                                   =30100×110030100×+×1100251001.2100+45100×2100                                                   =3030+30+90=30150=0.2                                                   

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Question 17:

A company has two plants to manufacture bicycles. The first plant manufactures 60% of the bicycles and the second plant 40%. Out of the 80% of the bicycles are rated of standard quality at the first plant and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be standard quality. Find the probability that it comes from the second plant.

Answer:

Let A, E1 and E2 denote the events that the cycle is of standard quality, plant I is chosen and plant II is chosen, respectively.

 PE1=60100     PE2=40100 Now, PA/E1=80100PA/E2=90100Using Bayes' theorem, we getRequired probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2+PE3PA/E3                                                   =40100×9010060100×80100+40100×90100                                                   =3648+36=3684=37                                                   

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Question 18:

Three urns A, B and C contain 6 red and 4 white; 2 red and 6 white; and 1 red and 5 white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the ball was drawn from urn A.

Answer:

Let A, E1 and E2 denote the events that the ball is red, bag A is chosen, bag B is chosen and bag C is chosen, respectively.

 PE1=13      PE2=13       PE3=13Now, PA/E1=610=35PA/E2=28=14PA/E3=16Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2+PE3PA/E3                                                   =13×3513×35+13×14+13×16                                                   =3535+14+16=3661                                                   

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Question 19:

In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?

Answer:

Let A, E1, E2 and E3 denote the events that the person suffers from the disease, is a smoker and a non-vegetarian, is a smoker and a vegetarian and the person is a non-smoker and a vegetarian, respectively.

 PE1=160400     PE2=100400       PE3=140400Now, PA/E1=35100PA/E2=20100PA/E3=10100Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2+ PE3PA/E3                                                   =160400×35100160400×35100+100400×20100+140400×10100                                                   =560560+200+140=560900=2845                                                   

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Question 20:

A factory has three machines A, B and C, which produce 100, 200 and 300 items of a particular type daily. The machines produce 2%, 3% and 5% defective items respectively. One day when the production was over, an item was picked up randomly and it was found to  be defective. Find the probability that it was produced by machine A.

Answer:

Let A, E1, E2 and E3 denote the events that the item is defective, machine A is chosen, machine B is chosen and machine C is chosen, respectively.

 PE1=100600     PE2=200600      PE3=300600Now, PA/E1=2100PA/E2=3100PA/E3=5100Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2+PE3PA/E3                                                   =100600×2100100600×2100+200600×3100+300600×5100                                                   =22+6+15=223                                                   

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Question 21:

A bag contains 1 white and 6 red balls, and a second bag contains 4 white and 3 red balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be white in colour. Find the probability that the drawn ball was from the first bag.

Answer:

Let A, E1 and E2 denote the events that the ball is white, bag I is chosen and bag II is chosen, respectively.

 PE1=12      PE2=12Now, PA/E1=17PA/E2=47Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =12×1712×17+12×47                                                   =11+4=15                                                   

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Question 22:

In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Further more, 60% of the students in the colleges are girls. A student selected at random from the college is found to be taller than 1.75 metres. Find the probability that the selected students is girl.

Answer:

Let A, E1 and E2 denote the events that the height of the student is more than 1.75 m, selected student is a girl and selected student is a boy, respectively.

 PE1=60100      PE2=40100Now, PA/E1=1100PA/E2=4100Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =60100×110060100×1100+40100×4100                                                   =66+16=622=311                                                   

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Question 23:

For A, B and C the chances of being selected as the manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8 and 0.5. If the change does take place, find the probability that it is due to the appointment of B or C.

Answer:

Let AE1E2 and E3 denote the events that the change takes place, A is selected, B is selected and C is selected, respectively.

 PE1=47     PE2=17      PE3=27Now,PA/E1=0.3PA/E2=0.8PA/E3=0.5Using Bayes' theorem, we get PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2+PE3PA/E3              =47×0.347×0.3+17×0.8+27×0.5             =1.21.2+0.8+1=1.23=1230=25 Required probability =1- PE1/A=1-25=35                                                   
 

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Question 24:

Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1 : 2 :4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3, respectively. If the change does not take place, find the probability that it is due to the appointment of C.

Answer:

Let E1, E2 and E3 be the events denoting the selection of A, B and C as managers, respectively.

PE1 = Probability of selection of A = 17

PE2 = Probability of selection of B = 27

PE3 = Probability of selection of C = 47

Let A be the event denoting the change not taking place.
PAE1 = Probability that A does not introduce change = 0.2
PAE2 = Probability that B does not introduce change = 0.5
PAE3 = Probability that C does not introduce change = 0.7
∴ Required probability = PE3A

By Bayes' theorem, we have

PE3A=PE3PAE3PE1PAE1+PE2PAE2+PE3PAE3= 47×0.717×0.2+27×0.5+47×0.7=2.80.2+1+2.8=2.84=0.7
 

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Question 26:

Of the students in a college, it is known that 60% reside in a hostel and 40% do not reside in  hostel. Previous year results report that 30% of students residing in hostel attain A grade and 20% of ones not residing in hostel attain A grade in their annual examination. At the end of the year, one students is chosen at random from the college and he has an A grade. What is the probability that the selected student is a hosteler?

Answer:

Let A, E1 and E2 denote the events that the selected student attains grade A, resides in a hostel and does not reside in a hostel, respectively.

 PE1=60100      PE2=40100Now, PA/E1=30100PA/E2=20100Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =60100×3010060100×30100+40100×20100                                                   =1818+8=913                                                   

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Question 27:

There are three coins. One is two headed coin, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer:

Let E1, E2 and E3 denote the events of choosing a two-headed coin, a biased coin and an unbiased coin, respectively.

Let A be the event that the coin shows heads.

 PE1=13     PE2=13     PE3=13Now, PA/E1=1PA/E2=75%=34PA/E2=12Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE2PA/E2                                                   =13×113×1+13×34+13×12                                                   =49                                                   

 

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Question 29:

Coloured balls are distributed in four boxes as shown in the following table:

Box Colour
Black White Red Blue
I
II
III
IV
3
2
1
4
4
2
2
3
5
2
3
1
6
2
1
5
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.

Answer:

Let A, E1, E2, E3 and  E4 denote the events that the ball is black, box I selected, box II selected, box III is selected and box IV is selected, respectively.

 PE1=14       PE2=14       PE3=14        PE4=14Now, PA/E1=318PA/E2=28PA/E3=17PA/E4=413 Using Bayes' theorem, we getRequired probability = PE3/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =14×1714×318+14×28+14×17+14×413                                                   =1716+14+17+413=156947 



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Question 28:

Assume that the chances of a patient having a heart attack is 40%. It is also assumed that meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options and patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Answer:

Let A, E1 and E2 denote the events that the selected person had a heart attack, did yoga and meditation, and followed the drug prescriptions, respectively.

 PE1=12          PE2=12Now,PA/E1=0.40×0.70=0.28PA/E2=0.40×0.75=0.30Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =12×0.2812×0.28+12×0.30                                                   =1429                                                   

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Question 30:

If a machine is correctly set up it produces 90% acceptable items. If it is incorrectly set up it produces only 40% acceptable item. Past experience shows that 80% of the setups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly set up.

Answer:

Let A be the event that the machine produces two acceptable items.
Also, let E1 represent the event that the machine is correctly set up and E2 represent the event that the machine is incorrectly set up

 PE1=0.8      PE2=0.2Now, PA/E1=0.9×0.9=0.81PA/E2=0.40×0.40=0.16Using Bayes' theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2                                                   =0.8×0.810.8×0.81+0.2×0.16                                                   =8185                                                   

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Question 31:

Bag A contains 3 red and 5 black balls, while bag B contains 4 red and 4 black balls. Two balls are transferred at random from bag A to bag B and then a ball is drawn from bag B at random. If the ball drawn from bag B is found to be red find the probability that two red balls were transferred from A to B.

Answer:

It is given that bag A contains 3 red and 5 black balls and bag B contains 4 red and 4 black balls.
Let E1, E2, E3 and A be the events as defined below:
E1