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Page No 296:

Question 1:

If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

Answer:

Number of letters in the word ALGORITHM = 9
Then total number of words in the above said letter of the word = 9!
If "GOR" remain together, then it is considered as 1 group.
Now Number of letters = 6 + 1 = 7
then number of word of GOR remain together = 7!

A1 L2 I3 T4 H5 M6 GOR7

The required probability=7!9!
=7!9×8×7!=172

Page No 296:

Question 2:

Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks?

Answer:

Let the couple occupied adjacent desks consider those two as 1
Now there are 4 + 1 = 5 persons to be assigned.
Now number of warp of assigning these five persons = 5!
Total number of warp of assigning 6 persons = 6!
So the probability tat the couple has adjacent desk=5!6!×2=13
Now as per the given conditions that probability that the married couple will have non-adjacent desks=1-13=23

Page No 296:

Question 3:

Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Answer:

Multiple of 2 from 1 to 1000 are 2, 4, 6, ..... 1000.

Let n be the number of terms of above series

Tn = a + (n – 1) d

1000 = 2 + (n – 1)2 ⇒ n = 500

So the number of multiple of 2 are 500.

Now the multiple of 9 are 9, 18, 27, ..... 999.

Tm = a + (m – 1)d

999 = 9 + (m – 1) 9 ⇒ m = 111

So the number of multiple of 9 are 111

Now the multiple of 2 and 9 both are 18, 36 54 ..... 999

So Tp = a + (p – 1)d

990 = 18 + (p – 1) 18 ⇒ p = 55

So the number of multiple of both 2 and 9 are 55.

So number of multiple of 2 or 9

= 500 + 111 – 55 = 556

∴ Required probability of multiples of 2 or 9
 

=5561000=0.556

Page No 296:

Question 4:

An experiment consists of rolling a die until a 2 appears.
(i) How many elements of the sample space correspond to the event that the 2 appears on the kth roll of the die?
(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the kth roll of the die?

Answer:

In a through a die there is 6 sample points
(i) If 2 appears on the kth roll of the die so first (k – 1) roll have 5 outcomes each and kth roll results 2 i.e. 1 outcome.
∴ Number of element of sample space correspond to the event that 2 appears on the kth roll of the die = 5k – 1.

(ii) If we consider that 2 appears not later than kth roll of the die, then it is possible that 2 comes in first thrown i.e. 1 outcome
If 2 does not appear in the first throw, then outcomes will be 5 and 2 comes in second thrown i.e. 1 outcome, possible outcome = 5 × 1 = 5
Simply, if 2 does not appear in second from and appears in third throw.

∴ Possible outcomes = 5 × 5 × 1

Then Series=1+5+5×5+5×5×5+.....+5k  1=1+5+52+53+.....+5k+1=15k-15-1=5k-14

Page No 296:

Question 8:

A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15, 0.20, 0.31, 0.26, .08. Find the probabilities that a particular surgery will be rated
(a) complex or very complex;
(b) neither very complex nor very simple;
(c) routine or complex
(d) routine or simple

Answer:

Let E1, E2, E3, E4 and E5 be the event that surgeries are rated as very complex, routine, simple or very simple respectively.

P(E1) = Very complex = 0.15
P(E2) = Complex = 0.2
P(E3) = routine = 0.31
P(E4) = simple = 0.26
P(E5) = Very Simple = 0.08

(a) P(complex or very complex)

= P(E2 or E1)
= P(E2∪E1) = P(E2) + P(E1) – P(E2∩E1)
= 0.20 + 0.15 – 0   [∵ P(E2∩E1) = 0 all events are independent]
= 0.35

(b) P(neither very complex nor very simple)
= P(E′1∩E′5) = P(E1∪E5)′
= 1 – P(E1∪E5)
= 1 – P(E1 + E5 – P(E1∩E5)
= 1 – [0.15 + 0.08 – 0]
= 1 – 0.15 – 0.08
= 1 – 0.23
= 1 – 0.77

(c) P(routine or complex) = P(E3∪E2)
                                     = P(E3) + P(E2)
                                     = 0.31 + 0.20 = 0.51

(d) P(routine or simple) = P(E3∪E4)
                                  = P(E3) + P(E4)
                                  = 0.31 + 0.26
                                  = 0.57



Page No 297:

Question 5:

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Answer:

It is given that 2x probability of even number = Probability of odd number

⇒ P(0) = 2P(E)

P(0) : P(E) = 2 : 1

Probability of occurring odd number=P0=22+1=23 and probability of occurring even number=PE=12+1=13

Now, G be the even that a number greater than 3 occurs in a single roll of die. So the possible out comes are 4, 5 and 6 out of which two are even and one odd.

Reqd. probability=PG=2×PE×P0=2×13×23=49

Page No 297:

Question 6:

In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

Answer:

Let E1 be the event that family own color television set and E2 be the event that family owns a black and white television set.
It is given that = P(E1) = 0.87

P(E2) = 0.36
P(E1∩E2) = 0.36

We have to find probability that a family owns either any one or both kind of sets i.e. P(E1∪E2)
Now by addition theorem
⇒ P(E1∪E2) = P(E1) + P(E2) – P(E1∩E2)
= 0.87 + 0.36 – 0.30
= 0.93

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Question 7:

If A and B are mutually exclusive events, P(A) = 0.35 and P(B) = 0.45, find
(a) P(AÍ´)
(b) P(BÍ´)
(c) P(A ⋃ B)
(d) P(A â‹‚ B)
(e) P(A â‹‚ BÍ´)
(f) P(AÍ´ â‹‚ BÍ´)

Answer:

Since it is given that A and B are mutually exclusive events.
∴ P(A â‹‚ B) = 0
and it is given P(A) = 0.35, P(B) = 0.45
(a) P(A′) = 1 – P(A) = 1 – 0.35 = 0.65
(b) P(B′) = 1 – P(B) = 1 – 0.45 = 0.55
(c) P(A⋃B) = P(A) + P(B) – P(Aâ‹‚B) = 0.35 + 0.45 – 0 = 0.80
(d) P(Aâ‹‚B) = 0
(e) P(Aâ‹‚B′) = P(A) – P(Aâ‹‚B)
                     = 0.35 – 0 = 0.35
(f) P(A′â‹‚B′) = (A⋃B)′ = 1 – P(A⋃B)
                                       = 1 – 0.8 = 0.2

Page No 297:

Question 9:

Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that
(a) C will be selected?
(b) A will not be selected?

Answer:

It is given that A is twice on likely to be selected as B.

P(A) = 2P(B)

PA2=PB

Which C is twice on likely to be selected as D

P(C) = 2P(D) ⇒ P(B) = 2P(D)

PA2=2PDPD=PA4

B and C are given about the same change of being selected

P(B) = P(C)

Now sum of probability = 1

P(A) = P(B) + P(C) + P(D) = 1

PA+PA2+PA2+PA4=14PA+2PA+2PA+PA=149PA=4PA=49

(a) P(C will be selected)=PC=PB=PA2=492=29

(b) P(A will be selected)=PA'=1-PA=1-49=59

Page No 297:

Question 10:

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}
You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John.
(a) Determine P (John promoted)

P (Rita promoted)
P (Aslam promoted)
P (Gurpreet promoted)
(b) If A = {John promoted or Gurpreet promoted}, find P (A).

Answer:

(a) Let E1 = John promoted

E2 = Rita promoted
E3 = Aslam promoted
E4 = Gurpreet promoted
Given, sample Space,

S = {John promoted, Rita promoted, Aslam, promoted, Gurpreet promoted}

i.e., S = {E1, E2, E3, E4}

It is given that, chances of John's promotion is same as that of Gurpreet

P(E1) = P(E4)

Rita's chances of promotion are twice as likely as John

P(E2) = 2P(E1)

and Aslam's chances of promotion are four times that of John

P(E3) = 4P(E1)

Now Sum of probability = 1

P(E1) + P(E2) + P(E3) + P(E4) = 1

P(E1) + 2P(E1) + 4P(E1) + P(E1) = 1

8P(E1) = 1

P(E1) =18

(b) A = John promoted or Gurpreet promoted
 
A = E1∪PE4

P(A) = P(E1∪E4) = P(E1) + P(E4) – P(E1∩E4)

=18+18-0=14       (Since P(E1∩E4) = 0 as they are independent



Page No 298:

Question 11:

The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections (for instance, P (A â‹‚ B) = .07). Determine

(a) P (A)
(b) P (B â‹‚ C )
(c) P (A â‹ƒ B)
(d) P (A â‹‚ B )
(e) P (B â‹‚ C)
(f) Probability of exactly one of the three occurs.

Answer:



(a) P(A) = 0.31 + 0.07 = 0.2

(b) P(B∩C¯) = P(B) – P(B∩C) = 0.07 + 0.1 + 0.15 – 0.15

= 0.17

(c) P(A∪B) = P(A) + P(B) – P(A∩B)
                     = 0.13 + 0.07 + 0.07 + 0.1 + 0.15 – 0.07
                     = 0.45

(d) P(A∩B¯) = P(A) – P(A∩B) = 0.13 + 0.07 – 0.07
                     = 0.13

(e) P(B∩C) = 0.15

(f) P(exactly one of three occurs) = 0.13 + 0.1 + 0.28
                                                      = 0.51

 

Page No 298:

Question 12:

One urn contains two black balls (labelled B1 and B2) and one white ball. A second urn contains one black ball and two white balls  (labelled W1 and W2). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.
(a) Write the sample space showing all possible outcomes
(b) What is the probability that two black balls are chosen?
(c) What is the probability that two balls of opposite colour are chosen?

Answer:

It is given that one of the two Urn is chosen, then a ball is randomly chosen from the Urn, then a second ball is chosen at random from the same Urn without replacement the first ball.



(a) Sample Space
S = {B1B2, B1W1, B2B1, B1W, WB1, WB2, BW1, BW2, W1B, W1W2, W2B, W2W1}
∴ Total sample point = 12

(b) If two black balls are chosen so the favourable events are B1B2, B2B1
i.e. 2
Required probability=212=16

(c) If two balls of opposite color are chosen so the favourable events are B1W1, B2W, WB1, WB2, BW1, BW2, W1B1, W2B i.e 8
Required probability=812=23

Page No 298:

Question 13:

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the Probability that
(a) All the three balls are white
(b) All the three balls are red
(c) One ball is red and two balls are white

Answer:

Number of red balls = 8 and number of white balls = 5
(a) P(all three balls are white)

=C35C313=5!3!2!13!10!3!=5!3!2!×10!3!13!=5×4×313×12×11=5143
(b) P(all the three balls are red)
=C38C313=8!5!3!13!10!3!=8!×10!×3!5!×3!×13!=8×7×613×12×11=28143
(c) P(one ball is red and balls are white)
=C18×C25C313=8×5!3!2!13!10!3!=8!×5!×10!×3!3!×2!×13!=8×1013×6×11=40143

Page No 298:

Question 14:

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that
(a) Four S’s come consecutively in the word
(b) Two I’s and two N’s come together
(c) All A’s are not coming together
(d) No two A’s are coming together.

Answer:

Total number of letters in word ASSASSINATION are 13
Out of which
A is three times
S is four times
I is 2 times
N is 2 times
T is 1 time
O is 1 time

(a) If four S come consecutively in the word then we considers these 4S as 1 group

Number of words when all S are together =10!3!2!2!1!
Total number of word using letters of word ASSASSINATION =13!3!4!2!2!1!1!
Required probability=10!3!2!2!13!3!4!2!2!=10!×3!4!2!2!3!2!2!×13!=4!13×12×11=241716=2143
 
(b) If 2I and 2N come together then there as 10 alphabets.
Number of word when 2I and 2N are come together=10!3!4!×4!2!2!
Required probability=10!×4!3!4!2!2!13!3!4!2!2!=10!4!×3!4!2!2!3!4!2!2!×13!=4!10!13!=2143

(c) If all A's are coming together, then there are 11 alphabets
Number of words when all A's come together =11!4!2!2!
Probability when all A's come together =11!4!2!2!13!4!3!2!2!=11!×4!3!2!2!4!2!2!13!=126
Required probability when all A's does not come together =1-126=2526

(d) If no two A's are together, then first we arrange the alphabets except A's
– S – S – S – S – I – N –  T – I – O –  N –
All the alphabets except A's are arranged in 10!4!2!2!
There are 11 vacant places between these alphabets
So 3A's can be placed in 11 places is C311 ways =11!3!8!
Total number of words when two A's together =11!3!8!×10!4!2!2!
Required probability=11!×10!3!8!4!2!×4!3!2!2!13!=10!8!×13×12=1526



Page No 299:

Question 15:

A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.

Answer:

Number of possible events = 52 and favourable events

= 4 king + 13 hearts + 26 red – 13 – 2
= 28
Required probability=2852=713

Page No 299:

Question 16:

A sample space consists of 9 elementary outcomes e1, e2, ..., e9 whose probabilities are
P(e1) = P(e2) = .08, P(e3) = P(e4) = P(e5) = .1
P(e6) = P(e7) = .2, P(e8) = P(e9) = .07
Suppose A = {e1, e5, e8}, B = {e2, e5, e8, e9}
(a) Calculate P (A), P (B), and P (A â‹‚ B)
(b) Using the addition law of probability, calculate P (A ⋃ B)
(c) List the composition of the event A ⋃ B, and calculate P (A ⋃ B) by adding the probabilities of the elementary outcomes.
(d) Calculate PB from P (B), also calculate PB directly from the elementary outcomes of B

Answer:

S = {e1, e2, e3, e4, e5, e6, e7, e8, e9}
A = {e1, e5, e8}
B = {e2, e5, e8, e9}
P(e1) = P(e2) = 0.08
P(e3) = P(e4) = P(e5) = 0.1
P(e6) = P(e7) = 2
P(e8) = P(e9) = 0.07

(a) P(A) = P(e1) + P(e5) + P(e8)

    = 0.08 + 0.1 + 0.07 = 0.25

(b) P(A⋃B) = P(A) + P(B)– P(A∩B)
Now, P(B) = P(e2) + P(e5) + P(e8) + P(e9)
= 0.08 + 0.1 + 0.07 + 0.07 = 0.32
P(A∩B) = (e5, e8) = 0.1 + 0.07 = 0.17
On substituting we have
P(A⋃B = P(A) + P(B) – P(A∩B)
= 0.25 + 0.32 – 0.17
= 0.40

(c) (A⋃B) = {e1, e2, e5, e8, e9}
P(A⋃B) = P(e1) + P(e2) + P(e5) + P(e8) + P(e9)
= 0.08 + 0.08 + 0.1 + 0.07 + 0.07 = 0.40

(d)  PB=1-PB
and B =e1 e3 e4 e6 e7PB=Pe1+Pe3+Pe4+Pe6+Pe7        =0.08+0.1+0.1+0.2+0.2=0.68

Page No 299:

Question 17:

Determine the probability p, for each of the following events.
(a) An odd number appears in a single toss of a fair die.
(b) At least one head appears in two tosses of a fair coin.
(c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.
(d) The sum of 6 appears in a single toss of a pair of fair dice.

Answer:

(a) When a die is throw the possible outcomes are

S = {1, 2, 3, 4, 5, 6} out of which 1, 3, 5 are odd.
Required probability=36=12

(b) When a fair coin is tossed two times the sample space is
S = {HH, HT, TH, TT}
At least one head, favourable events are HH, HT, TH.
Required probability=34

(c) Total cards = 52
favourable = 4 king + 2 of heart + 3 of spade
   = 4 + 1 + 1 = 6
Required probability=652=326

(d) When a pair of dice is rolled total sample parts are 36, out of which sum of six is
(1, 5) (5, 1) (2, 4) (4, 2) and (3, 3)
Required probability=536

Page No 299:

Question 18:

Choose the correct answer out of four given options:
In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is
(A) 17

(B) 27

(C) 37

(D) none of these

Answer:

In a non-leap year. There are 365 days which have 52 weeks and 1 day. It can be Tuesday or Wednesday. Then regarding probability of 53 Tuesday or 53 Wednesday will be 17.

Hence, the correct answer is option (A).

Page No 299:

Question 19:

Choose the correct answer out of four given options:
Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

(A) 186190

(B) 187190

(C) 188190

(D) 18C320

Answer:

Since the set of three consecutive numbers from 1 to 20 are
123, 234, 345, 456 ...... 18, 19, 20 . i.e., 18
Pnumbers are consecutive=18C320=181140=3190P numbers non consecutive=1-3190=187190

Hence, the correct answer is option (B).

Page No 299:

Question 20:

Choose the correct answer out of four given options:
While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours
(A) 2952

(B) 12

(C) 2651

(D) 2751

Answer:

Since in park of 52 cards 26 are red and 26 are black in colour.

P(both cards of opposite colour)=2652×2651+2652×2655=2×2652×2651=2651

Hence, the correct answer is option (C).



Page No 300:

Question 21:

Choose the correct answer out of four given options:
Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is

(A) 13

(B) 16

(C) 27

(D) 12

Answer:

If two persons sit next to each other, then consider two persons as 1 group. Now we have to arrange 6 persons
∴ Number of arrangement = 2! × 6!
Total number of arrangement of 7 persons = 7!
Required probability=2! 6!7!=27

Hence, the correct answer is option (C).

Page No 300:

Question 22:

Choose the correct answer out of four given options:
Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is

(A) 15

(B) 45

(C) 130

(D) 59

Answer:

We have, to form four four digit number using the digit 0, 2, 3 and 5 which are divisible by 5
If 0 is fixed at unit place = 3 × 2 × 1 = 6
If 5 is fixed at unit place = 2 × 2 × 1 = 4
Total form digit number divisible by 5 = 6 + 4 = 10
Regarding probability = 1018=59

Hence, the correct answer is option (D).

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Question 23:

Choose the correct answer out of four given options:
If A and B are mutually exclusive events, then
(A) P(A) ≤ P (B)
(B) P(A) ≥ P (B)
(C) P(A) < P (B)
(D) none of these

Answer:

For mutually exclusive events
PAB=0PAB=PA+PB-PABPAB=PA+PBPA+PB1PA+1-PB1PAPB

Hence, the correct answer is option (A).

Page No 300:

Question 24:

Choose the correct answer out of four given options:
If P (A⋃B) = P (A⋂B) for any two events A and B, then
(A) P(A) = P(B)
(B) P(A) > P(B)
(C) P(A) < P(B)
(D) none of these

Answer:

Given P(A⋃B) = P(A∩B)
⇒ P(A) + P(B) – P(A∩B) = P(A∩B
⇒ P(A) – P(A∩B) + P(B) – P(A∩B) = 0
But P(A) – P(A∩B) ≥ 0
and P(B) – P(A∩B) ≥ 0
⇒ P(A) – P(A∩B) = 0
and P(B) – P(A∩B) = 0
Since, sum of two non-negative numbers can be zero only when these numbers are zero
⇒ P(A) = P(A∩B)

P(B) = P(A∩B)
P(A) = P(B)

Hence, the correct answer is option (A).

Page No 300:

Question 25:

Choose the correct answer out of four given options:
6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is

(A) 1432

(B) 12431

(C) 1132

(D) none of these

Answer:

If all girls sit together, then considered it as 1 group
∴ Arrangement of 6 + 1 = 7 person in a row 57! and the girl interchanges their seats in 6! ways
Regarding probability=6!×7!12!=6!×7!12×11×10×9×8×7!=6×5×4×3×2×112×11×10×9×8=1132

Hence, the correct answer is option (C).

Page No 300:

Question 26:

Choose the correct answer out of four given options:
A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is

(A) 13

(B) 411

(C) 211

(D) 311

Answer:

Total number of alphabets in the word PROBABILITY = 11
Number of vowels = 4
Regarding probability P(vowel) =411

Hence, the correct answer is option (B).

Page No 300:

Question 27:

Choose the correct answer out of four given options:
If the probabilities for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is
(A) > . 5
(B) .5
(C) ≤ .5
(D) 0

Answer:

Given P(A fail) = 0.2

 P(B fail) = 0.3
P(either A or B fail) ≤ P(A fail) + P(B fail)
≤ 0.2 + 0.3
≤ 0.5

Hence, the correct answer is option (C).

Page No 300:

Question 28:

Choose the correct answer out of four given options:
The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then PA+PB is
(A) 0.4
(B) 0.8
(C) 1.2
(D) 1.6

Answer:

Given P(A⋃B) = 0.6 and P(A∩B) = 0.2
∴ P(A⋃B) = P(A) + P(B) – P(A∩B)

  0.6 = P(A) + P(B) – 0.2
P(A) + P(B) = 0.8

PA+PB=1-PA+1-PB=2-PA+PB=2-0.8=1.2

Hence, the correct answer is option (C).



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Question 29:

Choose the correct answer out of four given options:
If M and N are any two events, the probability that at least one of them occurs is
(A) P(M) + P(N) – 2P (Mâ‹‚N)
(B) P(M) + P(N) – P (Mâ‹‚N)
(C) P(M) + P(N) + P (Mâ‹‚N)
(D) P(M) + P(N) + 2P (Mâ‹‚N)

Answer:

If M and N are two events then P(M⋃N) = P(M) + P(N) – P(M∩N)

Hence, the correct answer is option (B).

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Question 30:

State whether the statement is True or False:
The probability that a person visiting a zoo will see the giraffee is 0.72, the probability that he will see the bears is 0.84 and the probability that he will see both is 0.52.

Answer:

P(to see girafee) = 0.72
P(to see bears) = 0.84
P(to see girafee and bear) = 0.52
∴ P(to see girafee or bear) = P(girafee) + P(bear) – P(girafee and bear)

   = 0.72 + 0.84 – 0.52 = 1.04
Which is not possible
Hence, the given statement is false.

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Question 31:

State whether the statement is True or False:
The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get compartment is 0.96.

Answer:

Let,
P(A) = Student will pass examination
P(B) = Student will get compartment
P(A) = 0.73 and P(A or B) = 0.96 P(B) = 0.13
P(A or B) = P(A) + P(B)

= 0.73 + 0.13 = 0.86
But it is given P(A or B) = 0.96
Hence, the given statement is false.

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Question 32:

State whether the statement is True or False:
The probabilities that a typist will make 0, 1, 2, 3, 4, 5 or more mistakes in typing a report are, respectively, 0.12, 0.25, 0.36, 0.14, 0.08, 0.11.

Answer:

Sum of these probability must equal to 1
P(0) + P(1) + P(2) + P(3) + P(4) + P(5)
= 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11
= 1.06 > 1      (greater than 1)
Hence, the given statement is false.

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Question 33:

State whether the statement is True or False:
If A and B are two candidates seeking admission in an engineering College. The probability that A is selected is .5 and the probability that both A and B are selected is at most .3. Is it possible that the probability of B getting selected is 0.7?

Answer:

As per given condition
P(A) = 0.5 and P(A∩B) ≤ 0.3
P(A) × P(B) ≤ 0.3
0.5 × P(B) ≤ 0.3

   P(B) ≤ 0.6
But it is given P(B) = 0.7
Hence, the given statement is false.

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Question 34:

State whether the statement is True or False:
The probability of intersection of two events A and B is always less than or equal to those favourable to the event A.

Answer:

Since P(A∩B) ≤ P(A)

Hence, the given statement is true.

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Question 35:

State whether the statement is True or False:
The probability of an occurrence of event A is .7 and that of the occurrence of event B is .3 and the probability of occurrence of both is .4.

Answer:

It is given P(A) = 0.7

 P(B) = 0.3
P(A∩B) = P(A) × P(B)
= 0.7 × 0.3
= 0.21
where as P(A∩B) = 0.4 (given)
Hence, the given statement is false.

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Question 36:

State whether the statement is True or False:
The sum of probabilities of two students getting distinction in their final examinations is 1.2.

Answer:

Since, these two events not related to the same sample space.
So, the sum of probability of two students getting distinction in their final examination may be 1.2.
Hence, the statement is true.

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Question 37:

Fill in the blank.
The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is _____.

Answer:

P(A) = P(winning) = 0.77
P(B) = P(tie) = 0.08
P(loosing game) = 1 – P(A) – P(B)

= 1 – 0.77 – 0.08
= 0.15

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Question 38:

Fill in the blank.
If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) = .1, P(e2) = .5, P(e3) = .1, then the probability of e4 is ______.

Answer:

Since sum of probability is equal to 1.

Pe1+Pe2+Pe3 + Pe4=10.1+0.5+0.1+Pe4=1Pe4=1-0.7=0.3Pe4=0.3

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Question 39:

Fill in the blank.
Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then E is _________.

Answer:

It is given S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}
E=S-E=2, 4, 6

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Question 40:

Fill in the blank.
If A and B are two events associated with a random experiment such that P( A) = 0.3, P(B) = 0.2 and P (A â‹‚ B) = 0.1, then the value of P (A â‹‚ B) is _______.

Answer:

It is known that
PAB=PA-PAB=0.3-0.1=0.2

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Question 41:

Fill in the blank.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is ________.

Answer:

Since,
PAB=PAB=1-PAB=1-PA+PB=1-0.5+0.3=1-0.8=0.2



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Question 42:

Match the proposed probability under Column C1 with the appropriate written description under column C2 :

C1 C2
Probability Written Description
(a) 0.95 (i) An incorrect assignment
(b) 0.02 (ii) No chance of happening
(c) – 0.3 (iii) As much chance of happening as not.
(d) 0.5 (iv) Very likely to happen
(e) 0  (v) Very little chance of happening

Answer:

(i) 0.95 is very likely is happen, so it is close to 1.
(ii) 0.02 is very little chance of happening as the probability is very low.
(iii) Probability of any event cannot be negative so is an incorrect assignment
(iv) 0.5 is as much chance of happening as not as it is midway between 0 and 1.
(v) 0 is no chance of happening.
Hence, (a)-(iv), (b)-(v), (c)-(i), (d)-(iii), (e)-(ii).

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Question 43:

Match the following

(a) If E1 and E2 are the two mutually exclusive events (i) E1 â‹‚ E2 = E1
(b) If E1 and E2 are the mutually exclusive and exhaustive events    (ii) (E1 – E2) ⋃ (E1 â‹‚ E2) = E1
(c) If E1 and E2 have common outcomes, then (iii) E1 ⋂ E2 = ϕ, E1 ⋃ E2 = S
(d) If E1 and E2 are two events such that E1 ⊂ E2 (iv) E1 â‹‚ E2 = Ï•

Answer:

(i) If E1 and Eare the two mutually exclusive events then E1 â‹‚ E2 = Ï•
(ii) If E1 and Eare the mutually exclusive and exhaustive events then E1 â‹‚ E2 = Ï• and E1 â‹ƒ E2 = S
(iii) If E1 and Ehave common outcomes, then (E1 – E2) ⋃ (E1 â‹‚ E2) = E1
(iv) If E1 and E2 are two events such that E1 ⊂ E2 ⇒ E1 â‹‚ E2 = E1
Hence, (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).



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