RD Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 14 Binomial Distribution are provided here with simple step-by-step explanations. These solutions for Binomial Distribution are extremely popular among class 12 Science students for Math Binomial Distribution 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 14 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 33.12:

Question 1:

There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.

Answer:

Let X denote the number of defective items in a sample of 8 items.  Then, X follows a binomial distribution with n = 8, p = (probability of getting a defective item) = 0.06 and q = 1-p=0.94

P(X=r)  =Cr8 (0.06)r (0.94)8-r, r=0,1,2,3,...8The required probability = Probability of not more than one defective item                                           = P (X1)                                             = P(X=0)+P(X=1)                                           =   C08 (0.06)0 (0.94)8-0+ C18 (0.06)1 (0.94)8-1                                            = (0.94)8 +8(0.06)(0.94)7                                           =  (0.94)7 0.94+0.48                                            = 1.42(0.94)7 



Page No 33.13:

Question 2:

A coin is tossed 5 times. What is the probability of getting at least 3 heads?

Answer:

Let X denote the number of heads in 5 tosses.X follows a binomial distribution with n =5; p = probability of getting a head = 12and q = 1-p =12P(X=r) = Cr512r125-r,  r =0,1,2...5The required probability = P(getting at least 3 heads)= P(X3) = P(X=3)+P(X=4)+P(X=5) = C35123125-3+ C45121125-1 + C55125125-0  
=10125 + 5125 + 1125 = 125(10+5+1)   = 125×16 = 12

Page No 33.13:

Question 3:

A coin is tossed 5 times. What is the probability that tail appears an odd number of times?

Answer:

Let X denote the number of tails when a coin is tossed 5 times.X follows a binomial distribution with n =5; p = 12;  q =1-p =12Then P(X=r) =Cr512r 12n-r  = Cr5125 The required probability = P(X=odd) =P(X=1)+P(X=3)+P(X=5)=C15125 +C35125 +C55125 = 125 [5+10+1] = 1632=12

Page No 33.13:

Question 4:

A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?

Answer:

Let X be the number of successes in 6 throws of the two dice.
 
Probability of success = Probability of getting a total of 9 
                                    = Probability of getting (3,6), (4,5), (5,4), (6,3) out of 36 outcomes

 p=436=19,   q =1-p =89and n = 6X follows a binomial distribution with n=6, p=19 and q=89P(X=r) = Cr619r  896-r The required probability = Probability of at least 5 successes= P(X5) = P(X=5)+P(X=6)=C56 195 896-5 +C66 196  896-6 = 6(8)+196=4996

Page No 33.13:

Question 5:

A fair coin is tossed 8 times, find the probability of 
  
    (i) exactly 5 heads       (ii)  at least six heads          (iii)   at most six heads.

Answer:


Let X denote the number of heads obtained when a fair is tossed 8 times.

Now, X is a binomial distribution with n = 8, p=12 and q=1-12=12.

PX=r=8Cr128-r12r=8Cr128, r=0, 1, 2,..., 8

(i) Probability of getting exactly 5 heads = PX=5=8C5128=56256=732

(ii) Probability of getting atleast 6 heads
=PX6=PX=6+PX=7+PX=8=8C6128+8C7128+8C8128=28+8+1×1256=37256
(iii) Probability of getting at most 6 heads
=PX6=1-PX=7+PX=8=1-8C7128+8C8128=1-8256+1256=1-9256=247256

Page No 33.13:

Question 6:

Find the probability of 4 turning up at least once in two tosses of a fair die.

Answer:

Let X be the probability of getting 4 in two tosses of a fair die.
X follows a binomial distribution with n =2; p =16 and q =56;
P(X=r) =  Cr216r562-rProbability of getting 4 at least once = P(X1)  = 1-P(X=0) = 1- C02160562-0 = 1-2536= 1136

Page No 33.13:

Question 7:

A coin is tossed 5 times. What is the probability that head appears an even number of times?

Answer:

Let X be the number of heads that appear when a coin is tossed 5 times.

X follows a binomial distribution with n =5 and p= q=12
P(X=r) = Cr512r125-r               =   Cr5125

P (head appears an even number of times)=P(X=0)+P(X=2)+P(X=4)=  C05125+C25125+C45125  = 1+10+525= 1632=12

Page No 33.13:

Question 8:

The probability of a man hitting a target is 1/4. If he fires 7 times, what is the probability of his hitting the target at least twice?

Answer:

Let X be number of times the target is hit. Then, X follows a binomial distribution with n =7, p =14and q =34

P(X=r) = Cr714r347-rP( hitting the target at least twice)=P(X2)   =  1-P(X=0) +P(X=1)=  1-C07140347-0-C17141347-1  = 1-347 - 714346=  1-116384(2187+5103)   =  1-36458192=45478192

Page No 33.13:

Question 9:

Assume that on an average one telephone number out of 15 called between 2 P.M. and 3 P.M. on week days is busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?

Answer:

Let X be the number of busy calls for 6 randomly selected telephone numbers.

X follows a binomial distribution with n =6 ;p = one out of 15 = 115and q = 1415
P(X=r) = Cr6115r14156-rProbability that at least 3 of them are busy = P(X3) = 1-{P(X=0)+P(X=1)+P(X=2)}= 1 -C06115014156-0+C16115114156-1+C26115214156-2                 

=1-14156+61514155+11514154

Page No 33.13:

Question 10:

If getting 5 or 6 in a throw of an unbiased die is a success and the random variable X denotes the number of successes in six throws of the die, find P (X ≥ 4).

Answer:

Let X denote the number of successes, i.e. of getting 5 or 6 in a throw of die in 6 throws.

Then, X follows a binomial distribution with n =6;
p = of getting 5 or 6 = 16+16=13; q = 1-p = 23;P(X=r) = Cr613r236-rP(X 4) = P(X=4)+P(X=5)+P(X=6)=C46134236-4+C56135236-5+C66136236-6 = 136(60+12+1) =73729         

Page No 33.13:

Question 11:

Eight coins are thrown simultaneously. Find the chance of obtaining at least six heads.

Answer:

Let X be the number of heads in tossing 8 coins.
X follows a binomial distribution with n =8; p = 12and q = 12;
P(X=r) = Cr812r128-r  = Cr8128Probability of obtaining at least 6 heads = P(X6) = P(X=6)+P(X=7)+P(X=8)=    C68128 + C78128 + C88128  = 12828+8+1 = 37256

Page No 33.13:

Question 12:

Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade?

Answer:

Let X denote the number of spade cards when 5 cards are drawn with replacement.  Because it is with replacement,

X follows a binomial distribution with n = 5; p =1352=14;    q = 1-p = 34

P(X=r) =Cr514r345-r

(i) P(All cards are spades)= P(X=5) =C55145340  = 11024(ii) P(only 3 cards are spades) = P(X=3) =C35143342  = 1102490 = 45512iii P(none is spade)= P(X=0)=C05140345  = 2431024

Page No 33.13:

Question 13:

A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
(i) none is white?
(ii) all are white?
(iii) any two are white?

Answer:

Let X be the number of white balls drawn when 4 balls are drawn with replacement
 
X follows binomial distribution with n = 4.

p = Probability for a white ball = No of white ballsTotal no. of balls = 520=14and q = 1-p = 34P(X=r) = Cr414r344-r(i) Prob that none is white = P(X=0) =C04140344-0= 81256ii Prob that all are white = P(X=4) = C44144344-4= 1256iii Prob that any two  are white = P(X=2) =  C24142344-2= 54256=27128

Page No 33.13:

Question 14:

A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.

Answer:

Let X be the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.
Then, X follows a binomial distribution with n =5;

p = Probability of getting a ticket bearing number divisible by 10.    p = 1100(10) = 110;  q=910;P(X=r) = Cr5110r9105-rProbability that all the tickets bear numbers divisible by 10= P(X=5) =C5511059105-5=11059100=1105 

Hence, required probability is 1105

Page No 33.13:

Question 15:

A bag contains 10 balls, each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Answer:

Let X be the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

As this is with replacement, X follows a binomial distribution with n = 4;
p = probabilty that a ball randomly drawn bears digit 0 =110;q=1-p = 910;

P(X=r) = Cr4110r9104-r  P(none bears the digit 0) = P(X=0) = C0411009104-0   = 9104

Page No 33.13:

Question 16:

In a large bulk of items, 5 per cent of the items are defective. What is the probability that a sample of 10 items will include not more than one defective item?

Answer:

Let X denote the number of defective items in a sample of 10 items.

X follows a binomial distribution with n =10; p = probability of defective items = 5% = 0.05 ;q = 1-p =0.95

P(X=r) =Cr10(0.05)r(0.95)10-rProbability ( sample of 10 items will include not more than one defective item)= P(X1) = P(X=0)+P(X=1) =C010(0.05)0(0.95)10-0+C110(0.05)1(0.95)10-1   =( 0.95)9(0.95+0.5)= 1.45(0.95)9   

=192092920;      ( 1.45 = 2920and 0.95 = 1920)

Page No 33.13:

Question 17:

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none will fuse after 150 days of use
(ii) not more than one will fuse after 150 days of use
(iii) more than one will fuse after 150 days of use
(iv) at least one will fuse after 150 days of use

Answer:

Let X be the number of bulbs that fuse after 150 days.

X follows a binomial distribution with n = 5, p =0.05 and q =0.95
Or p = 120and q = 1920P(X=r) = Cr5120r19205-r(i) Probability (none will fuse after 150 days of use ) = P(X=0)  =C05120019205-0 =19205(ii)  Probability (not more than 1 will fuse after 150 days of use) = P(X1)  = P(X=0)+P(X=1) =19205+5C1120119205-1 = 192041920+520  = 6519204


(iii) Probability more than one will fuse after 150 days of use = P(X>1) = 1-P(X1)= 1-6519204         P(X1)=6519204    iv Probability  at least one will fuse after 150 days of use = P(X1)= 1-P(X=0) = 1-19205          P(X=0=19205  

Page No 33.13:

Question 18:

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Answer:

Let X be the number of people that are right-handed in the sample of 10 people.

X follows a binomial distribution with n = 10,
p =90% =0.9  and q = 1-p = 0.1P(X=r) = Cr10(0.9)r(0.1)10-rProbability that at most 6 are right-handed = P(X6)= P(X=0)+ P(X=1)+ P(X=2)+ P(X=3)+ P(X=4)+ P(X=5)+ P(X=6)= 1-{P(X=7)+ P(X=8)+ P(X=9)+ P(X=10)}= 1- r=710Cr10(0.9)r(0.1)10-r   

Page No 33.13:

Question 19:

A bag contains 7 green, 4 white and 5 red balls. If four balls are drawn one by one with replacement, what is the probability that one is red?

Answer:

Let X denote the number of red balls drawn from 16 balls with replacement.
X follows a binomial distribution with n = 4,p = 516,  q =1-p = 1116

P(X=r) =  Cr4516r11164-rP(One ball is red)=P(X=1) = C14516111164-1= 451611163=5411163

Page No 33.13:

Question 20:

A bag contains 2 white, 3 red and 4 blue balls. Two balls are drawn at random from the bag. If X denotes the number of white balls among the two balls drawn, describe the probability distribution of X.

Answer:

Let X denote the number of white balls when 2 balls are drawn from the bag.

X follows a distribution with values 0,1 or 2. 


P(X=0) = P(All balls non-white) = C27C29= 4272=2136P(X=1) = P( Istball white and IInd ball non-white )                  =C17C12C29= 1436P(X=2) = P(Both balls white) = C22C29= 136It can be shown in tabular form as follows.X                     0           1          2P(X)             2136      1436        136

Page No 33.13:

Question 21:

An urn contains four white and three red balls. Find the probability distribution of the number of red balls in three draws with replacement from the urn.

Answer:

As three balls are drawn with replacement, the number of white balls, say X, follows binomial distribution with n =3
p =37 and  q =47P(X=r) = Cr337r473-r,  r=0,1,2,3P(X=0) =C03370473-0  P(X=1) = C13371473-1  P(X=2) = C23372473-2  P(X=3) =C33373473-3X        0         1            2          3P(X)  64343  144343    108343   27343

Page No 33.13:

Question 22:

Find the probability distribution of the number of doublets in 4 throws of a pair of dice.

Answer:

Let X be the number of doublets in 4 throws of a pair of dice.

X follows a binomial distribution with n =4,
p = No of getting (1,1)(2,2)...(6,6) = 636=16q =1-p = 56P(X=r) = Cr416r564-r, r= 0,1,2,3,4P(X=0) = C04160564-0P(X=1) = C14161564-1P(X=2) = C24162564-2P(X=3) = C34163564-3P(X=4) = C44164564-4The distribution is as follows.X           0                    1                    2                  3              4  P(X)    6251296        5001296       1501296          201296    11296

Page No 33.13:

Question 23:

Find the probability distribution of the number of sixes in three tosses of a die.

Answer:

Let X be the number of 6 in 3 tosses of a die.

Then X follows a binomial distribution with n =3.

p = 16,  q=1-p =56P(X=r) = Cr316r563-r, r=0,1,2,3P(X=0) = C03160563-0P(X=1) = C13161563-1P(X=2) = C23162563-2P(X=3) = C33163563-3

Hence, the distribution of X is as follows.X             0         1       2        3P(X)    125216 75216  15216   1216

Page No 33.13:

Question 28:

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the mean and variance of number of red cards. 

Answer:


It is given that the cards are drawn successively with replacement so the the events are independent. Therefore, the drawing of the cards follow binomial distribution.

​Probability of drawing a red card = p2652=12

q = 1 − p1-12=12

Also, n = 3

Let X be the random variable denoting the number of red cards drawn from a well shuffled pack of 52 cards.

PX=r=Cr3123-r12r=Cr3123,  r=0, 1, 2, 3

Probability of drawing no red ball = PX=0=C03123=18
Probability of drawing one red ball = PX=1=C13123=38
Probability of drawing two red balls = PX=2=C23123=38
Probability of drawing three red balls = PX=3=C33123=18

Thus, the probability distribution of X is as follows:

xi pi pixi pixi2
0 18 0 0
1 38 38 38
2 38 68 128
3 18 38 98
    pixi=128 pixi2=3

Mean of =pixi=128=32

Variance of =pixi2-Mean2=3-94=34



Page No 33.14:

Question 24:

A coin is tossed 5 times. If X is the number of heads observed, find the probability distribution of X.

Answer:

Let X = number of heads in 5 tosses. Then the binomial distribution for X has n =5, p =12 and  q =12
P(X=r) = Cr512r125-r, r = 0,1,2,3,4,5               = Cr525Substituting r =0,1,2,3,4,5 we get the following probability disrtribution.X           0         1      2      3       4        5P(X)   132    532 1032 1032 532   132

Page No 33.14:

Question 25:

An unbiased die is thrown twice. A success is getting a number greater than 4. Find the probability distribution of the number of successes.

Answer:

Let X denote getting a number greater than 4 .

Then,  X follows a binomial distribution with n=2

  p = P(X>4)=P(X=5 or 6)= 16+16=13q =1-p = 23P(X=r) = Cr2 13r232-r  , r=0,1,2   Substituting for r we get probability distribution of X as follows.

X            0      1       2P(X)    49   49   19

Page No 33.14:

Question 26:

A man wins a rupee for head and loses a rupee for tail when a coin is tossed. Suppose that he tosses once and quits if he wins but tries once more if he loses on the first toss. Find the probability distribution of the number of rupees the man wins.

Answer:

Let X be the number of rupees the man wins.

First, let us assume that he gets head in the first toss.Probability would be  12.  Also, he wins Re.1.The second possibility is that he gets a tail in the first toss.  Then he tosses again.Suppose he gets head in the second toss.Then, he wins Re 1 in the second toss but loses Re 1 in the first toss.  So, the money won  =Rs. 0Probability for winning Rs .0 =Getting tail  in Ist toss  x  Getting head in IInd toss=12×12=14The third possibility is getting tail in the  Ist toss and also tail in the IInd  toss.Then,  the money that he would win = -2       (As he loses Rs. 2)Probability for the third possibility =  12×12=14Probability distribution of X is as follows.X           1         0        -2P(X)    12       1 4        14

Page No 33.14:

Question 27:

Five dice are thrown simultaneously. If the occurrence of 3, 4 or 5 in a single die is considered a success, find the probability of at least 3 successes.

Answer:

Let X denote the occurrence of 3,4 or 5 in a single die. Then, X follows binomial distribution with 
n=5.
Let p=probability of getting 3,4, or 5 in a single die .

 p = 36=12

q=1-12=12 P(X=r) = Cr512r125-rP(at least 3 successes) =P(X3) = P(X=3)+P(X=4)+P(X=5)= C35123125-3+C45124125-4+C55125125-5= C35+C45+C5525= 12

Page No 33.14:

Question 28:

The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is 28×96108.

Answer:

Let X denote the number of defective items in the items produced by the company.
Then, X follows binomial distribution with n = 8.

p = 10%=110q= 1-p = 910Hence, the distribution is given byP(X=r) = Cr8110r9108-rProb of getting 2 defective items =P(X=2) =C2811029108-2   = 28 x 96108

Page No 33.14:

Question 29:

A card is drawn and replaced in an ordinary pack of 52 cards. How many times must a card be drawn so that (i) there is at least an even chance of drawing a heart (ii) the probability of drawing a heart is greater than 3/4?

Answer:

(i) Let p denote the probability of drawing a heart from a deck of 52 cards. So,

 p=1352=14and q=1-q=1-14 = 34

Let the card be drawn n times. So, binomial distribution is given by: P(X=r)=Crnprqn-r
Let X denote the number of hearts drawn from a pack of 52 cards.
We have to find the smallest value of n for which P(X=0) is less than 14
P(X=0) < 14
C0n14034n-0<1434n<14Put n=1, 341 not less than 14      n=2, 342 not less than 14       n=3, 343 not less than 14So, smallest value of n=3
Therefore card must be drawn three times.

(ii) Given the probability of drawing a heart > 34
1 - P(X=0) > 34
1- C0n14034n-0>341-34n>341-34>34n14>34n

For n=1, 341 not less than 14 n=2, 342 not less than 14 n=3, 343 not less than 14 n=4, 344 not less than 14 n=5, 345 not less than 14
So, card must be drawn 5 times.

Page No 33.14:

Question 30:

The mathematics department has 8 graduate assistants who are assigned to the same office. Each assistant is just as likely to study at home as in office. How many desks must there be in the office so that each assistant has a desk at least 90% of the time?

Answer:

Let k be the number of desks and X be the number of graduate assistants in the office.
therefore, X=8, p=12, q=12
According to the given condition,
 PXk>90%PXk>0.90PX>k<0.10P(X=k+1, k+2,....8)<0.10
Therefore, P(X > 6) = P(X=7 or X=8)
C78128 + C88128 =0.04

Now, P(X > 5) = P(X = 6, X = 7 or X = 8) = 0.15
P(X > 6)  < 0.10
        
So, if there are 6 desks then there is at least 90% chance for every graduate to get a desk.
 

Page No 33.14:

Question 31:

An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 6 heads.

Answer:

Let X be the number of heads in tossing the coin 8 times.
X follows a binomial distribution with n =8
p =12 and q=12Hence, the distribution is given by P(X=r) = Cr812r128-r, r=0,1,2,3,4,5,6,7,8Required probability =P(X6) = P(X=6)+P(X=7)+P(X=8)=  C68+ C78+ C8828= 28+8+1256=37256

Page No 33.14:

Question 32:

Six coins are tossed simultaneously. Find the probability of getting
(i) 3 heads
(ii) no heads
(iii) at least one head

Answer:

Let X denote the number of heads obtained in tossing 6 coins.
Then, X follows a binomial distribution with n=6,

p = 12 and q=12Hence, the distribution is given byP(X=r) =Cr612r126-r  , r=0,1,2,3,4,5,6= Cr626i P(getting 3heads) =P(X=3) =C3626= 2064=516ii P(getting no head) =P(X=0) =C0626= 126  = 164iii P(getting at least 1 head) = P(X1)  =1- P(X=0)  = 1-164=6364

Page No 33.14:

Question 33:

Suppose that a radio tube inserted into a certain type of set has probability 0.2 of functioning more than 500 hours. If we test 4 tubes at random what is the probability that exactly three of these tubes function for more than 500 hours?

Answer:

Let X denote the number of tubes that function for more than 500 hours. 
Then, X follows a binomial distribution with n =4.
Let p be the probability that the tubes function more than 500 hours.

Here, p=0.2, q =0.8Hence, the distribution is given byP(X=r) =Cr4(0.2)r(0.8)4-r, r= 0,1,2,3,4Therefore, required probability = P(X=3) = 4(0.2)3(0.8) = 0.0256

Page No 33.14:

Question 34:

The probability that a certain kind of component will survive a given shock test is 34. Find the probability that among 5 components tested
(i) exactly 2 will survive
(ii) at most 3 will survive

Answer:

Let X denote the number of components that survive shock.
Then, X follows  a binomial distribution with n = 5.
Let p be the probability that a certain kind of component will survive a given shock test.
p = 34and q=14Hence, the disrtibution is given byP(X=r) = Cr534r145-r, r=0,1,2,3,4,5i P(exactly 2 will survive)=P(X=2)  =C25342145-2  = 10×91024= 0.0879

ii P(atmost 3 will survive) = P(X3)= P(X=0)+ P(X=1)+ P(X=2) +P(X=3)=C05340145-0+C15341145-1+C25342145-2+C35343145-3= 145+534144+10342143+10343142= 1+15+90+2701024= 3761024= 0.3672

Page No 33.14:

Question 35:

Assume that the probability that a bomb dropped from an aeroplane will strike a certain target is 0.2. If 6 bombs are dropped, find the probability that
(i) exactly 2 will strike the target
(ii) at least 2 will strike the target

Answer:

Let X be the number of bombs that hit the target.
Then, X follows a binomial distribution with n = 6
Let p be the probability that a bomb dropped from an aeroplane will strike the target.

 p =0.2 and q =0.8Hence, the distribution is given byP(X=r) =Cr60.2r0.86-r(i) P(exactly 2 will strike the target)=P(X=2) = C26(0.2)2(0.8)4=15(0.04)0.4096=0.2458(ii) P(at least 2 will strike the target) = P(X2)  = 1-[P(X=0)+P(X=1)]= 1- (0.8)6-6(0.2)(0.8)5 = 1-0.2621-0.3932= 0.3447

Page No 33.14:

Question 36:

It is known that 60% of mice inoculated with a serum are protected from a certain disease. If 5 mice are inoculated, find the probability that
(i) none contract the disease
(ii) more than 3 contract the disease

Answer:

Let X be the number of mice that contract the disease .
Then, X follows a binomial distribution with n =5.
Let p be the probability of mice that contract the disease.
 p=0.4 and q = 0.6Hence, the distribution is given byP(X=r)=Cr50.4r0.65-r  , r=0,1,2,3,4,5i P(X=0) =C050.400.65-0= (0.6)5 = 0.0778ii P(X>3) =P(X=4)+P(X=5)=C450.440.65-4+C550.450.65-5=0.0768+0.01024= 0.08704

Page No 33.14:

Question 37:

An experiment succeeds twice as often as it fails. Find the probability that in the next 6 trials there will be at least 4 successes.

Answer:

Let X denote the number of successes in 6 trials.

It is given that successes are twice the failures. p=2q p+q=13q=1q=13 p=1-13=23 n =6Hence, the distribution is given by

P(X=r)=Cr623r136-r , r=0,1,2.....6P(atleast 4 successes)=P(X4) = P(X=4)+P(X=5)+P(X=6)C46234136-4+C56235136-5+C66236136-6= 15(24)+6(32)+6436= 240+192+64729= 496729

Page No 33.14:

Question 38:

In a hospital, there are 20 kidney dialysis machines and the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.

Answer:

Let X denote the number of machines out of service during a day.Then, X follows a binomial distribution with n=20 Let p be the probability of any machine out of service during a day.p= 0.02 and q =0.98   Hence, the distribution is given byP(X=r) =Cr20 0.02r0.9820-r  , r=0,1,2.....20P(exactlly 3 machines will be out of the service on the same day)=P(X=3) =C3 200.0230.9820-3 = 1140(0.000008)(0.7093)=  0.006469

Page No 33.14:

Question 39:

The probability that a student entering a university will graduate is 0.4. Find the probability that out of 3 students of the university
(i) none will graduate
(ii) only one will graduate
(iii) all will graduate

Answer:

Let X be the number of students that graduate from among 3 students.
Let p=probability that a student entering a university  will graduate.

Here , n =3, p=0.4 and q = 0.6
Hence, the distribution is given by
P(X=r)=Cr30.4r0.63-r   , r=0,1,2,3 

i P(X=0) = q3  =0.216ii P(X=1) = 3(0.4)(0.36) =0.432iii P(X =3) = p3 = 0.064

Page No 33.14:

Question 40:

Ten eggs are drawn successively, with replacement, from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.

Answer:

Let X be the number of defective eggs drawn from 10 eggs.
Then, X follows a binomial distribution with n =10
Let p be the probability that a drawn egg is defective.

 p =10%=110 , q =910Hence, the distribution is given by P(X=r) =Cr10110r91010-r   , r=0,1,2....10P(there is at least one defective egg) =P(X1) = 1-P(X=0) =1-C010110091010-0= 1-91010 

=1-9101010

Page No 33.14:

Question 41:

In a 20-question true-false examination, suppose a student tosses a fair coin to determine his answer to each question. For every head, he answers 'true' and for every tail, he answers 'false'. Find the probability that he answers at least 12 questions correctly.

Answer:

Let X denote the number of correct answers.
Then, X follows a binomial distribution with n =20

Let p be the probability of a correct answer. p= getting a head and a right answer to be true or getting a tail and a right answer to be falsep= 12 q=1- 12=12Hence, the distribution is given byP(X=r) = Cr2012r1220-r   , r=0,1,2,3......20= Cr20220Probability that the student answers at least 12 questions correctly=P(X12) = P(X=12)+P(X=13)+...+P(X=20)= C1220+C1320+...+C2020220



Page No 33.15:

Question 42:

Suppose X has a binomial distribution with n = 6 and p=12. Show that X = 3 is the most likely outcome.

Answer:

We have n=6 and p=12 q = 1-p =12Hence, the distribution is given byP(X=r) =Cr612r126-r, r=0,1,2,3,4,5,6= Cr6126       

P(X=r) = Cr626 By substituting r =0,1,2,3,4,5 and 6, we get the following distribution for X.X                0      1      2        3       4       5       6P(X)       164 664 1564   2064  1564  664  164

Comparing the probabilities, we get that X = 3 is the most likely outcome.

Page No 33.15:

Question 43:

In a multiple-choice examination with three possible answers for each of the five questions out of which only one is correct, what is the probability that a candidate would get four or more correct answers just by guessing?

Answer:

Let X be the number of right answers in the 5 questions. 
X can take values 0,1,2...5.
X follows a binomial distribution with n =5
.p =probability of guessing right answer = 13 q =probability of guessing wrong answer = 23Hence, the distribution is given byP(X=r)=Cr5 13 r 235-r   , r=0,1,2,...5 P(The student guesses 4 or more correct answers) = P(X4) = P(X=4)+P(X=5)=C45134231+C55135230  = 10+135=11243

Page No 33.15:

Question 44:

A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1100. What is the probability that he will win a prize
(i) at least once
(ii) exactly once
(iii) at least twice

Answer:

Let X denote the number of times the person wins the lottery. 
Then, X follows a binomial distribution with n = 50.

Let p be the probability of winning a prize. p = 1100, q=1-1100=99100Hence, the distribution is given byP(X=r)  = Cr501100r9910050-r , r=0,1,2 ...50i P(winning at least once) = P(X0) = 1-P(X-0)= 1-9910050 
ii P(winning exactly once) = P(X=1) =C150110019910050-1   =  129910049 

iii  P(winning at lease twice) = P(X2) = 1-P(X=0)-P(X=1)= 1-9910050-C150 ×1100×9910049 =1-9949 ×14910050

Page No 33.15:

Question 45:

The probability of a shooter hitting a target is 34. How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?

Answer:

Let the shooter fire n times and let X denote the number of times the shooter hits the target.
Then, X follows binomial distribution with p =34and q =14 such that

PX=r=Crn 34r14n-r PX=r=Crn 3r4nIt is given that PX1 >0.991-PX=0 >0.991-14n>0.9914n<0.014n>10.014n>100The least value of n satisfying this inequality is 4. Hence, the shooter must fire at least 4 times.

Page No 33.15:

Question 46:

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90% ?

Answer:

Suppose the man tosses a fair coin n times and X denotes the number of heads in n tosses.
As p=12 and  q=12, P(X=r) =Crn12r12n-r, r=0,1,2,3....nIt is given that PX1 >0.91-PX=0 >0.91-C0n 12n>0.912n<1102n>10n=4,5,6....Hence, the man must toss the coin at least 4 times. 

Page No 33.15:

Question 47:

How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%?

Answer:

Let X be the number of heads and n be the minimum number of times that a man must toss a fair coin so that probability of X1 is more than 80% and X follows a binomial distribution with

 p = 12,  q=12P(X=r) = Crn12nWe have P(X1) = 1-P(X=0) =1-C0n12n =1-12nand P(X1) >80%1-12n>80% = 0.8012n<1-0.80 =0.20 2n>10.2 =5;  We know, 22  <5 while 23>5So, n =3 So, n should be atleast 3.

Page No 33.15:

Question 48:

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.

Answer:

Let p denote the probability of getting a doublet in a single throw of a pair of dice. Then,
p = 636 = 16q = 1-p=1-16=56

p= Let X be the number of getting doublets in 4 throws of a pair of dice. Then, X follows a binomial distribution with n =4,
p=16and q=56P(X=r)=Probability of getting r doubletsP(X=r) = Cr4(16)r(56)4-r;  r= 0,1,2,3,4If X=0,  then P(X=0) = C04(16)0(56)4-rP = 564If X=1, then P = C14161564-1=23563If X=2, then  P = C24162564-2=16562 If X=3, then P =  C34163564-3=103163 

If X=4, then  P =C44(16)4(56)4-4  =164= 11296



X                0              1                  2             3            4P(X)         (56)4    23(56)3         25216       5324      11296

Page No 33.15:

Question 49:

From a lot of 30 bulbs that includes 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Answer:

Let X denote the number of defective bulbs in a sample of 4 bulbs drawn successively with replacement .
Then, X follows a binomial distribution with the following parameters: n=4, p =630=15and q =45
Then, the distribution is given byP(X=r) =Cr415r454-r  , r=0,1,2,3,4

 P X=0=454                   = 256625 P X=1  = 4151453                    = 256625 PX=2= 6152452                 = 96625 PX=3 = 4153451                  = 16625 PX=4  =154                  = 1625



X         0      1     2     3     4

P(X)    256625 256625 96625 16625 1625

Page No 33.15:

Question 50:

Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws.
                                                                                                                                                                                           [NCERT EXEMPLAR]

Answer:

Let getting a multiple of 3 be a success.We have,p=the probability of getting a success=26=13So, q=the probability of getting a failure=1-p=1-13=23Let X denote the number of success in a sample of 10 trials. Then,X follows binomial distribution with parameters n=10, p=13 and q=23 PX=r=Cr10prq10-r=Cr1013r2310-r=Cr101310210-r=Cr10210-r310, r=0,1,2,...,10Now,Required probability=PX8=PX=8+PX=9+PX=10=C810210-8310+C910210-9310+C1010210-10310=45×22310+10×2310+1310=180+20+1310=201310

Page No 33.15:

Question 51:

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.                              [NCERT EXEMPLAR]

Answer:

Let getting an odd number be a success in a trial.

We have,

p=probability of getting an odd number in a trial=36=12Also, q=1-p=1-12=12Let X denote the number of success in a sample of 5 trials. Then,X follows binomial distribution with parameters n=5 and p=q=12 PX=r=Cr5prq5-r=Cr512r125-r=Cr5125, where r=0,1,2,3,4,5Now,Required probability=PX=3=C35125=1032=516

Page No 33.15:

Question 52:

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
                                                                                                                                                                                         [NCERT EXEMPLAR]

Answer:

Let hitting the target be a success in a shoot.

We have,

p=probability of hitting the target=0.25=14Also, q=1-p=1-14=34Let X denote the number of success in a sample of 7 trials. Then,X follows binomial distribution with parameters n=7 and p=14 PX=r=Cr7prq7-r=Cr714r347-r=Cr737-r47, where r=0,1,2,3,4,5Now,Required probability=PX2=1-PX=0+PX=1=1-C073747+C173647=1-218716384+510316384=1-729016384=909416384=45478192

Page No 33.15:

Question 53:

A factory produces bulbs. The probability that one bulb is defective is 150 and they are packed in boxes of 10. From a single box, find the probability that
(i) none of the bulbs is defective
(ii) exactly two bulbs are defective
(iii) more than 8 bulbs work properly                                                                                                                             [NCERT EXEMPLAR]

Answer:

Let getting a defective bulb from a single box is a success.

We have,

p=probability of getting a defective bulb=150Also, q=1-p=1-150=4950Let X denote the number of success in a sample of 10 trials. Then,X follows binomial distribution with parameters n=10 and p=150 PX=r=Cr10prq10-r=Cr10150r495010-r=Cr104910-r5010, where r=0,1,2,3,...,10Now,i Required probability=Pnone of the bulb is defective=PX=0=C01049105010=49105010=495010ii Required probability=Pexactly two bulbs are defective=PX=2=C2104985010=45×4985010iii Required probability=Pmore than 8 bulbs work properly=Patmost one bulb is defective=PX0=PX=0+PX=1=C01049105010+C1104995010=49105010+10×4995010=499501049+10=594995010

Page No 33.15:

Question 54:

A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.

Answer:


Let p denote the probability of drawing a defective pen. Then,
p=220=110q=1-p=1-110=910
Let X denote the number of defective pens drawn. Then, X is a binomial variate with parameter n = 5 and p=110.
Now, P(X = r) = Probability of drawing r defective pens = Cr5110r9105-r, r = 0, 1, 2, 3, 4, 5
∴ Probability of drawing at most 2 defective pens
= P(X  ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
=C0511009105+C1511019104+C2511029103=910381100+5×9100+10100=7291000×136100=0.99144 
 



Page No 33.25:

Question 1:

Can the mean of a binomial distribution be less than its variance?

Answer:

No.

The mean of a binomial distribution is np and variance is npq.

If mean is less than its variance, then np <npq

As both n and p are positive, we can divide both sides by np.

We get 1<q, which is not true as q <1 under all circumstances. (As p+q=1, q cannot be greater than 1)

So, the mean of a binomial distribution cannot be less than its variance.

Page No 33.25:

Question 2:

Determine the binomial distribution whose mean is 9 and variance 9/4.

Answer:

It is given that mean, i.e. np = 9 and variance, i.e. npq =94

npqnp =94×19   = 14  Hence, q=14  and p =1-q = 34When p = 34,np = 3n4 9=3n4n =12P(X=r) =Cr1234r1412-r, r=0,1,2,3,4,5,.....12

Page No 33.25:

Question 3:

If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.

Answer:

Given:  Mean = 9 and variance = 6

 np =9       ...(1)  npq =6      ...(2) Dividing eq (2) by eq (1), we get q =23and p =1-q=13As np =9, substituting the value of p, we get n3=9  or n =27P(X=r) = Cr2713r2327-r, r=0,1,2....27

Page No 33.25:

Question 4:

Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.

Answer:

Number of trials in the binomial distribution = 5

If p is the probability for success, then

 np + npq = 4.8  
Or 5p+5p (1-p) = 4.8

5p + 5p-5p2  =4.8  Or  p2-2p+0.96 =0By factorising, we get  (p-0.8)(p-1.2) =0As p cannot exceed 1, p =0.8 or 45and q= 1-p = 15 P(X=r) = Cr545r155-r, r=0,1,2,....5

Page No 33.25:

Question 5:

Determine the binomial distribution whose mean is 20 and variance 16.

Answer:

Mean, i.e. np =20          ....(1)

Variance, i.e. npq =16       ....(2)

Dividing eq (2) by eq (1), we get npqnp= 1620q = 45 p = 1-q  p= 15  As np = 20   n = 100P(X=r) =Cr10015r45100-r , r=0,1,2....100

Page No 33.25:

Question 6:

In a binomial distribution the sum and product of the mean and the variance are 253 and 503 respectively. Find the distribution.

Answer:

 Given:Sum of the mean and variance= 253np+npq=253   np 1+q=253        ...(1)Product of the mean and variance = 503 np(npq) = 503       ...(2)Dividing eq (2) by eq (1), we get  np(npq)np(1+q)= 503×325npq1+q=2npq  = 2(1+q)  np(1-p) =  2(2-p)np = 2(2-p)(1-p) Substituting this value in np+npq =253, we get2(2-p)(1-p) (2-p) = 253  6(4-4p+p2)  = 25-25p6p2+p-1 =0 (3p-1)(2p+1) =0   p =13or -12. As p cannot be negative, the only answer for p is 13.
q=1-p =23  np+npq = 253n131+23 =253n = 15P(X=r)  = Cr1513r2315-r, r=0,1,2......15

Page No 33.25:

Question 7:

The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.

Answer:

Given that mean, i.e. np = 20       ...(1)
and standard deviation, i.e. npq = 4

npq  =4    npq =16      ...(2)Dividing eq (2) by eq (1), we get  q= 1620=45and p = 15;    n = Meanp= 100 P(X=r) = Cr10015r45100-r,  r=0,1,2.....100Therefore, the parameters are n=100 and p = 15

Page No 33.25:

Question 8:

If the probability of a defective bolt is 0.1, find the (i) mean and (ii) standard deviation for the distribution of bolts in a total of 400 bolts.

Answer:

Total number of bolts (n) = 400 and p = prob of defective bolts = 0.1

(i) Mean = np = 400(0.1) =40

(ii) Variance = npq = 40(1-0.1) = 36

So, the standard deviation = 36  =6

Page No 33.25:

Question 9:

Find the binomial distribution whose mean is 5 and variance 103.

Answer:

Mean of binomial distribution, i.e. np =5

Variance, i.e. npq =103
q= VarianceMean= 23and p = 1-q= 13np = 5  n= 15P(X=r) = Cr1513r2315-r,  r=0,1,2......15

Page No 33.25:

Question 10:

If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.

Answer:

Total number of ships (n)  = 500
Let X denote the number of ships returning safely to the ports.
p=910 and q =1-p =110

Mean = np = 450 and Variance =npq = 45  Mean =450 Standard deviation = 45  = 6.71

Page No 33.25:

Question 11:

The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P (X = 0), P (X = 1) and P (X ≥ 2).

Answer:

Given: mean =16 and variance = 8
Let n and p be the parameters of the distribution.
 
That is, np = 16 and npq  = 8

q= npqnp=12and p = 1-q =12 np = 16 n = 32P(X=r) = Cr3212r1232-r, r=0,1,2....32,P(X=0) =1232 P(X=1) = 321232=122   P(X2) = 1-P(X=0)-P(X=1)=   1-1232-1227= 1-1+32232    = 1-33232

Page No 33.25:

Question 12:

In eight throws of a die, 5 or 6 is considered a success. Find the mean number of successes and the standard deviation.

Answer:

Let X denote the number of successes in 8 throws.

n =8 
p = probability of getting 5 or 6 =26=13 and q = 23

Mean (np) =83Variance (npq) = 169Standard deviation= Variance = 43So, mean = 2.66 and standard deviation = 1.33

Page No 33.25:

Question 13:

Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.

Answer:

Here, n =8
Let p be the probability of number of boys in the family.

p =12, q=12Expected number of boys =mean  np = 4

Page No 33.25:

Question 14:

The probability that an item produced by a factory is defective is 0.02. A shipment of 10,000 items is sent to its warehouse. Find the expected number of defective items and the standard deviation.

Answer:

Here, n =10,000
Let p (the probability of getting a defective item) = 0.02
q =1-0.02 = 0.98

Mean= Expected number of defective items np =200Variance (npq) = 200(0.98)                            = 196Standard deviation = Variance  = 14So, mean =200 and standard deviation =14

Page No 33.25:

Question 15:

A dice is thrown thrice. A success is 1 or 6 in a throw. Find the mean and variance of the number of successes.

Answer:

Here,  n =3
 p = probability of getting 1 or 6 =13and q =1-13=23Mean = np =1Variance = npq = 23

Page No 33.25:

Question 16:

If a random variable X follows a binomial distribution with mean 3 and variance 3/2, find P (X ≤ 5).

Answer:

Mean(np) =3 and variance (npq)=32 q=12and p =1-12n=Meanp n=6

Hence, the distribution is given by P(X=r) = Cr612r126-r  , r=0,1,2...6=  Cr626 P(X5) = 1-P(X=6)  = 1-164=6364

Page No 33.25:

Question 17:

If X follows a binomial distribution with mean 4 and variance 2, find P (X ≥ 5).

Answer:

Here, mean (np) = 4
variance (npq ) =2

q= 12and p =12   n=Mean p=4×2=8
Hence, the distribution is given by P(X=r) =Cr812r128-r, r=0,1,2......8P(X5) = P(X=5)+P(X=6)+P(X=7)+P(X=8)=128C58+C68+C78+C88   = (56+28+8+1)28= 93256

Page No 33.25:

Question 18:

The mean and variance of a binomial distribution are 43 and 89 respectively. Find P (X ≥ 1).

Answer:

Mean np =43and Variance npq = 89q = 23and  p =1-23= 13Therefore, n = Meanp = 4
Hence, the distribution is given by P(X=r) = Cr413r234-r, r=0,1,2,3,4P(X1) = 1-[P(X=0)]  = 1-234  = 81-1681= 6581

Page No 33.25:

Question 19:

If the sum of the mean and variance of a binomial distribution for 6 trials is 103, find the distribution.

Answer:

Given that n = 6 
The sum of mean and variance of a binomial distribution for 6 trials is 103.
6p+6pq = 10318p+18p(1-p) =10 18p2-36p+10 =0(3p-1)(6p-10) =0   p =13or 53p=53          (Neglected as it is greater than 1) p=13q= 1-p = 23Hence, the distribution is given by P(X=r) = Cr613r236-r, r=0,1,2.....6

Page No 33.25:

Question 20:

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes and, hence, find its mean.

Answer:

Let X be the number of times a doublet is obtained in four throws.

Then, p = probability of success in one throw of a pair of dice =636=16
and q = 56;  n = 4P(X=r) = Cr416r564-r, r= 0,1,2,3,4As n =4 and p = 16, mean = np = 46= 23
 P(X=r) = Cr456r16n-r, r=0,1,2,3,4The distribution is as follows:X             0           1            2             3        4P(X)     564    2064     15064     50064   164

Page No 33.25:

Question 21:

Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.

Answer:

Total number of outcomes when two dice are thrown = 6 × 6, i.e. 36

Let X be the number of doublets in three throws of a pair of dice.
Then, X follows a binomial distribution with n = 3
p = P(Getting a doublet in three throws) =636=16and q = 56P(X=r) =Cr316r563-r, r=0,1,2,3Mean np = 316  = 12       
The distribution is as follows:X              0                     1                        2                          3P(X)     563   3161563-1   3162563-2     163              125216           75216                  15216                 1216

Page No 33.25:

Question 22:

From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.                                                                                            [CBSE 2014]

Answer:

Let getting a defective bulb in a trial be a success.

We have,

p=probability of getting a defective bulb=515=13 andq=probability of getting non-defective bulb=1-p=1-13=23Let X denote the number of success in a sample of 4 trials. Then,X follows binomial distribution with parameters n=4 and p=13 PX=r=Cr4prq4-r=Cr413r234-r=Cr424-r34, where r=0,1,2,3,4i.e.PX=0=C042434=1681,PX=1=C142334=3281,PX=2=C242234=2481,PX=3=C342134=881,PX=4=C442034=181

So, the probability distribution of X is given as follows:
 

X: 0 1 2 3 4
P(X): 1681 3281 2481 881 181

Now,

Mean, EX=0×1681+1×3281+2×2481+3×881+4×181=32+48+24+481=10881=43Note: We can also calculate the mean of the binomial distribution byMean, EX=np=4×13=43

Page No 33.25:

Question 23:

A die is thrown three times. Let X be 'the number of twos seen'. Find the expectation of X.                                          [NCERT EXEMPLAR]

Answer:

We have,p=probability of getting a number two in a throw=16 andq=1-p=1-16=56As, X denote 'the number of twos seen'.So, X follows binomial distribution with parameters n=3 and p=16 EX=np=3×16=12

Page No 33.25:

Question 24:

A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes.       [NCERT EXEMPLAR]

Answer:

We have,p=probability of getting an even number on a toss=36=12 andq=1-p=1-12=12Let X denote a success of getting an even number on a toss. Then,X follows binomial distribution with parameters n=2 and p=12 VarX=npq=2×12×12=12



Page No 33.26:

Question 25:

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribtution.                                                                                                                              [CBSE 2015]

Answer:

We have,p=probability of getting a spade in a draw=1352=14 andq=1-p=1-14=34Let X denote a success of getting a spade in a throw. Then,X follows binomial distribution with parameters n=3 and p=14 PX=r=Cr3prq3-r=Cr314r343-r=Cr333-r43=2764Cr33r, where, r=0,1,2,3So, the probability distibution of X is given by:PX=r=2764Cr33r, where, r=0,1,2,3Now,Mean, EX=np=3×14=34

Page No 33.26:

Question 26:

An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find mean and variance of the distribution.

Answer:

Let X denote the total number of red balls when four balls are drawn one by one with replacement.
P (getting a red ball in one draw) = 23 
P (getting a white ball in one draw) = 13

0 1 2 3 4
P(X) 134 23133 . C14 232132 . C24 23313 . C34 234
  181 881 2481 3281 1681

Using the formula for mean, we have
X = PiXi
Mean (X¯) = 0×181+1 881+22481+3 3281+4 1681=1818+48+96+64=21681=83

Using the formula for variance, we have
Var(X)= PiXi2 -PiXi2
Var (X) = 0×181+1 881+42481+9 3281+16 1681-832=64881-649=89
Hence, the mean of the distribution is 83 and the variance of the distribution is 89.

Page No 33.26:

Question 27:

Five bad oranges are accidently mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.

Answer:

Let X be the random variable denoting the number of bad oranges drawn.

P (getting a good orange) = 2025=45 

P (getting a bad orange) = 525=15

The probability distribution of X is given by
 

0 1 2 3 4
P(X) 454= 256625 C1445315 =256625  C24452152 =96625 C3445153 = 16625 154 = 1625

Mean of X is given by

X = PiXi
= 0×256625+1×256625+2×96625+3×16625+4×1625=1625256+192+48+4=45

Variance of X is given by

Var(X)= PiXi2 -PiXi2
=0×256625+1×256625+4×96625+9×16625+16×1625-452=1625256+384+144+16-1625=800625-1625=400625=1625

Thus, the mean and vairance of the distribution are 45 and 1625, respectively.



Page No 33.27:

Question 1:

In a binomial distribution, if n = 20 and q = 0.75, then write its mean.

Answer:

n= 20 , q =0.75

p = 1-q = 0.25Mean = np = 20(0.25) =5Thus, mean =5

Page No 33.27:

Question 2:

If in a binomial distribution mean is 5 and variance is 4, write the number of trials.

Answer:


Mean =5 and Variance= 4np =5 and npq =4    q= 0.8  p =1-q =0.2& np = n(0.2) =5 (given)n = 50.2= 25

Page No 33.27:

Question 3:

In a group of 200 items, if the probability of getting a defective item is 0.2, write the mean of the distribution.

Answer:


It is given that the binomial distribution's p =0.2 and number of items (n) = 200

Hence, mean, i.e. np = 200 (0.2) =40

Page No 33.27:

Question 4:

If the mean of a binomial distribution is 20 and its standard deviation is 4, find p.

Answer:

Standard deviation of the binomial distribution =4 Variance=Standard deviation2Variance, i.e. npq= 16 Mean =np = 20q = 1620=45and p = 1-q = 15

Page No 33.27:

Question 5:

The mean of a binomial distribution is 10 and its standard deviation is 2; write the value of q.

Answer:


Mean of the binomial distribution, i.e. np =10Variance = Standard deviation2, i.e. npq =4q= VarianceMean =0.4   

Page No 33.27:

Question 6:

If the mean and variance of a random variable X with a binomial distribution are 4 and 2 respectively, find P (X = 1).

Answer:

Given, np=4 and npq=2p = 1-VarianceMean         = 1-24        =12and q=12 and n = npp       =412       = 8Hence, the binomial distribution is given byPX=r=Cr8 12r128-r , r=0,1,2,3.......8P(X=1) = 8128                      = 132

Page No 33.27:

Question 7:

If the mean and variance of a binomial variate X are 2 and 1 respectively, find P (X > 1).

Answer:


Mean =2 ,Variance =1 q=VarianceMean= 12and p =1-12=12n = Meanp= 212=4The binomial distribution is given byP(X=r)=Cr412r124-r P(X=0)=C04120124-0  ,   r=0,1,2,3,4=124P(X>1) = 1-P(X=0)  =  1-124  = 1516

Page No 33.27:

Question 8:

If in a binomial distribution n = 4 and P (X = 0) = 1681, find q.

Answer:

In the given binomial distribution, n = 4 and
P(X=0) = 1681  Binomial distribution is given byP(X=0)=C04 p0q4-0=q4  We know that P(X=0) = 1681  q4  =1681 q4=234q =23

Page No 33.27:

Question 9:

If the mean and variance of a binomial distribution are 4 and 3, respectively, find the probability of no success.

Answer:

Mean (np)= 4  

Variance (npq )= 3

.q  =  34  Hence, p =1-34 = 14  and n = Meanp= 4×4 =16Therefore, the binomial distribution is given byP(X=r) =Cr16 14r3416-r     , r=0,1,2....rProbability of no success = C0161403416-0=3416

Page No 33.27:

Question 10:

If for a binomial distribution P (X = 1) = P (X = 2) = α, write P (X = 4) in terms of α.

Answer:

For binomial distribution of X,P(X=r)= Crn(p)r(q)n-r,    r= 0,1,2,..., nP(X=1)=np(q)n-1   P(X=2) =C2np2 (q)n-2np(q)n-1=C2np2 (q)n-2 =α  Simplifying the above equation we get,q=n-12p 2q = np-p   On putting, q =1-p we get    2-2p = np-p   p(n+1) =2                   ..... (i)Also, P(X=1) = αnp(1-p)n-1 =α         .....(ii)

Note: We cannot find the value of n as (i) and (ii) are not linear and hence we cannot find the value of P(X = 4)

Page No 33.27:

Question 11:

An unbiased coin is tossed 4 times. Find the mean and variance of the number of heads obtained.                                              [CBSE 2015]

Answer:

We have,p=probability of getting a head in a toss=12,q=probability of getting a tail in a toss=12Let X denote a success of getting a head in a toss. Then,X follows binomial distribution with parameters n=4 and p=12 Mean, EX=np=4×12=2Also, variance, VarX=npq=4×12×12=1

Page No 33.27:

Question 12:

If X follows binomial distribution with parameters n = 5, p and P(X = 2) = 9P(X = 3), then find the value of p.                           [CBSE 2015]

Answer:

We have,X follows binomial distribution with parameters n=5, p and PX=2=9PX=3.So, PX=r=Cr5prq5-r, where r=0,1,2,3,4,5 and q=1-pAs, PX=2=9PX=3C25p2q3=9C35p3q210p2q3=9×10p3q2q=9p1-p=9p        As, q=1-p10p=1 p=110

Page No 33.27:

Question 1:

In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective?
(a) 9105

(b) 910

(c) 10−5

(d) 122

Answer:

(a) 9105

Let X denote the number of defective bulbs.

Hence, the binomial distribution is given by

n =5 , p = 10100=110 & q = 90100=910Hence, the distribution is given by P(X=r)=Cr5110r9105-rP(X=0)= 9105



Page No 33.28:

Question 2:

If in a binomial distribution n = 4, P (X = 0) = 1681, then P (X = 4) equals
(a) 116

(b) 181

(c) 127

(d) 18

Answer:

(b) 181

In the given binomial distribution, n = 4 and
P(X=0) = 1681  Binomial distribution is given byP(X=0)=C04 p0q4-0=q4  We know that P(X=0) = 1681  q4  =1681 q4=234q =23 p=1-23=13Then , P(X=4)=C44 p4q4-4=134=181

Page No 33.28:

Question 3:

A rifleman is firing at a distant target and has only 10% chance of hitting it. The least number of rounds he must fire in order to have more than 50% chance of hitting it at least once is
(a) 11
(b) 9
(c) 7
(d) 5

Answer:

(c) 7

Let p=chance of hitting a distant target
p =10% or p= 0.1

q =1-0.1= 0.9Let n  be the least number of rounds .P(hitting atleast once) = P(X1)  1-P(X=0) 50% 1-P(X=0)  0.5P(X=0) 0.5(0.9)n0.5Taking log on both the sides, we get n log 0.9  log 0.5  nlog 0.5log 0.9n7.2 Therefore, 7 is the least number of rounds that he must fire in order to have more than 50% chance of hitting the target at least once. 

Page No 33.28:

Question 4:

A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is
(a) 15/28
(b) 2/15
(c) 15/213
(d) None of these

Answer:

(c) 15/213

Let X denote the number of heads in a fixed number of tosses of a coin .Then, X is a binomial variate
with parameters  n and p=12

Given that P (X=7) =P (X = 9).  Also, p = q = 0.5

P(X=r)  = Crn(0.5)r(0.5)n-r =Crn (0.5)nP(X=7) =C7n(0.5)n   and P(X=9)=C9n(0.5)nIt is given that P(X=7)=P(X=9) C7n(0.5)n =C9n(0.5)nn!7! n-7 !=n!9! n-9 !9×8=n-7n-8n2-8n-7n+56=72n2-15n-16=0n+1n-16=0n=-1 or n=16 n=- 1   (Not possible as n denotes the number of tosses of a coin) n=16 Hence, P(X=2) =C216(0.5)16  = 16.152×1216= 15213

Page No 33.28:

Question 5:

A fair coin is tossed 100 times. The probability of getting tails an odd number of times is
(a) 1/2
(b) 1/8
(c) 3/8
(d) None of these

Answer:

(a) 1/2

Here n=100
Let X denote the number of times a tail is obtained.

Here, p =q = 12P(X=odd) = P(X=1,3,5,....99)  = C1100+C3100+.....+C9910012100= Sum of odd coefficients in binomial expansion in (1+x)10012100= 2(100-1)2100= 12

Page No 33.28:

Question 6:

A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is
(a) C1020×56620

(b) 120×57610

(c) 84×56610

(d) None of these

Answer:

(c) 84×56610


Let p be the probabilty of obtaining a six in a single throw of the die .Then ,p =16and q =1-16=56Obtaining a fourth six in the tenth throw of the die means that in the first nine throws there are 3 sixes and the fourth six is obtained in the tenth throw. Therefore, required probability=P(Getting 3 sixes in the first nine throws) P(Getting a six in the tenth throw)=C39 p3q9-3 p=C39163566×16= 84 x 56610

Page No 33.28:

Question 7:

If X is a binomial variate with parameters n and p, where 0 < p < 1 such that PX=rPX=n-ris independent of n and r, then p equals
(a) 1/2
(b) 1/3
(c) 1/4
(d) None of these

Answer:

(a) 1/2

Given that P(X=r) = k P(X=n-r), where k is independent of n and r .

Crn prqn-r =   k Cn-rn pn-rqrWe have Crn=Cn-rn  and also q=1-pHence, the equation changes to the following:pr(1-p)n-r  = k pn-r(1-p)r(1-p)n-2r  = k pn-2rqpn-2r=k    This is possible when p =q and k becomes 1.Hence, p = q = 12

Page No 33.28:

Question 8:

Let X denote the number of times heads occur in n tosses of a fair coin. If P (X = 4), P (X = 5) and P (X = 6) are in AP, the value of n is
(a) 7, 14
(b) 10, 14
(c) 12, 7
(d) 14, 12

Answer:

(a) 7, 14

Here, p=12and q=12Binomial distribution is given by P(X=r)=Crn12r12n-r

P (X = 4), P (X = 5), P(X = 6) are in A.P.

C4n+C6n=2C5nn(n-1)(n-2)(n-3)24!+n(n-1)(n-2)(n-3)(n-4)(n-5)26!= n(n-1)(n-2)(n-3)(n-4)5!By simplifying, we get12+(n-4)(n-5)2(30)= n-45Taking LCMas 60, we get30+n2-9n+20 = 12n-48 n2-21n+98 =0 (n-7)(n-14) =0n =7, 14

Page No 33.28:

Question 9:

One hundred identical coins, each with probability p of showing heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, the value of p is
(a) 1/2
(b) 51/101
(c) 49/101
(d) None of these

Answer:

(b) 51/101

 Let X denote the number of coins showing head. Therefore, X follows a binomial distribution with p and n as parameters.Given that P(X=50)=P(X=51)C50100p50q50 = C51100p51q49on simplifying we get,5150=pq5150=p1-p      (since p+q=1)p=51101

Page No 33.28:

Question 10:

A fair coin is tossed 99 times. If X is the number of times head appears, then P (X = r) is maximum when r is
(a) 49, 50
(b) 50, 51
(c) 51, 52
(d) None of these

Answer:

(a) 49, 50

When a coin is tossed 99 times, the number of heads X follows a binomial distribution with
p = q=12= 0.5P(X=r) =Crn(0.5)r(0.5)n-r=Crn(0.5)nAs (0.5)n is common to all r it is enough if we find the maximum of Crn.We know that for odd number of n, there will be two equal maximum terms,i.e. when r=n-12and r = n+12Hence, n =99 So, the maximum is obtained when r = 49 or 50

Page No 33.28:

Question 11:

The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8, is
(a) 7
(b) 6
(c) 5
(d) 3

Answer:

(d) 3

Let X denote the number of coins. 
Then, X follows a binomial distribution with

p = 12 ,  q = 12It is given that  P(X1) 0.81-P(X=0) 0.8P(X=0)1-0.8   P(X=0)=0.212n0.2    2n10.22n5This is possible when n3So, the least value of n is 3.

Page No 33.28:

Question 12:

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is
(a) 2/3
(b) 4/5
(c) 7/8
(d) 15/16

Answer:

(d) 15/16

Mean =2 and variance =1

np =2 and npq = 1q=12 p = 1-12=12 n=Meanpn =4Hence, the distribution is given byPX=r=Cr412r124-r  , r=0,1,2,3,4P(X1) = 1-P(X=0) = 1-124=1516

Page No 33.28:

Question 13:

A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p equals
(a) 1/3
(b) 2/3
(c) 2/5
(d) 3/5

Answer:

(a) 1/3

Probability of heads = p
Probability of Tails = (1 − p)
Probability that first head appears at even turn
Pe = (1 − p)p + (1 − p)3p+(1 − p)5p + .....
     = (1 − p)p (1 + (1 − p)2 + (1 − p)4+ .....)

     =1-pp11-1-p2=1-pp1-p2+2p

Pe=1-p2-pPe=251-p2-p=255-5p=4-2p3p=1p=13

Page No 33.28:

Question 14:

If X follows a binomial distribution with parameters n = 8 and p = 1/2, then P (|X − 4| ≤ 2) equals
(a) 118128

(b) 119128

(c) 117128

(d) None of these

Answer:

(b) 119128

n=8,   p = 12 q=1-12=12Hence, the distribution is given byP(X=r) = Cr812r128-r        PX-42  = P(-2X-42) =P(2X6)=P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=C28+C38+C48+C58+C6828= 28+56+70+56+28256=238256=119128

Page No 33.28:

Question 15:

If X follows a binomial distribution with parameters n = 100 and p = 1/3, then P (X = r) is maximum when r =
(a) 32
(b) 34
(c) 33
(d) 31

Answer:

(c) 33

The binomial distribution is given by,
fr, n, p=PX=r=crn pr 1-pn-r
This value can be maximum at a particular r, which can be determined as follows,
fr+1, n, pfr, n, p=1cr+1npr+1×1-pn-r-1crnpr×1-pn-r=n-rpr+1 1-p=1
On substituting the values of n = 100, p=13, we get
100-r×13r+1 1-13=1100-r13=r+123100-r=r+12100-r=2r+298=3r3r=98r=983
The integer value of r satisfies (n + 1)p − 1 ≤ m < (n + 1)p

f (r, n, p) is montonically increasing for r < m and montonically decreasing for r > m

as983m<1013

∴ The integer value of r is 33.


 



Page No 33.29:

Question 16:

A fair die is tossed eight times. The probability that a third six is observed in the eighth throw is
(a) C27×5567

(b) C27×5568

(c) C27×5566

(d) None of these

Answer:

(b) C27×5568


Let p be the probabilty of obtaining a six in a single throw of the die .Then ,p =16and q =1-16=56Obtaining a third six in the eighth throw of the die means that in first seven throws there are 2 sixes and the third six is obtained in the eighth throw . Therefore,required probability=P(Getting 2 sixes in the first seven throws) P(Getting six in the eighth throw)=C27 p2q7-2 p=C27162565 ×16= C27 x 5568

Page No 33.29:

Question 17:

Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9 is
(a) 357

(b) 1157

(c) 8157

(d) None of these

Answer:

Answer: (d)

Let p= probability that a selected coupon bears number 9.
p=915=35
n = number of coupons drawn with replacement
X = number of coupons bearing number 9
Probability that the largest number on the selected coupons does not exceed 9
= probability that all the coupons bear number 9
= P(X=7) = C77p7q0=377
Similarly, probability that largest number on the selected coupon bears the number 8 will be

P(X=7) = C77p7q0=8157         (since, p will become 815)
Hence required probability will be = 377-8157
So, option (d)



 

 

Page No 33.29:

Question 18:

A five-digit number is written down at random. The probability that the number is divisible by 5, and no two consecutive digits are identical, is
(a) 15

(b) 159103

(c) 354

(d) None of these

Answer:

Let number be abcde
Case 1 : e = 0
a, b, c can be filled in 9 × 9 × 9 ways
c = 0 ⇒ 9 × 8 × 1 ways and d has 9 choices
c ≠ 0 ⇒ (9 × 9 × 9 – 9 × 8 × 1) = 657
in the case d has 8 choices ⇒ 657 × 8
Total case = 9 × 8 × 1 × 9 + 657 × 8 ⇒ 5904
Case 2 : e = 5
If c = 5,
if a ≠ 5 then a, b, c can be filled in 8 × 8 × 1 = 64 ways
if a = 5 then a, b, c can be filled in 1 × 9 × 1 = 9 ways
if c ≠ 5, then first 3 digits can be filled in 729 – 64 – 9 = 656 ways
here d has 8 choices
No. of member ending in 5 and no two consecutive digits being identical ​⇒ (64 + 9) × 9 + 656 × 8

⇒ 5905
​Total cases ⇒ 5904 + 5905 ⇒ 11809
 
​​​Required Probability=1180990000

Hence, None of these

Page No 33.29:

Question 19:

A coin is tossed 10 times. The probability of getting exactly six heads is
(a) 512513

(b) 105512

(c) 100153

(d) C610

Answer:

(b) 105512

Let X denote the number of heads obtained in 10 tosses of a coin . Then, X follows a binomial distribution with n =6 , p=12=qThe distribution is given by P(X=r)=Cr1012r1210-r P(X=6) = C610210=  10529=105512

Page No 33.29:

Question 20:

If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is
(a) C6161410346

(b) C6161463410

(c) C612120346

(d) C612146346

Answer:

(b) C6161463410

Mean (np) = 4 and Variance (npq) = 3  

 q=34p =1-34=14and n = 16Let X denotes the number of successes in 16 trials .Then, P(X=r) =Cr1614r3416-rP(X=6) = Probability (getting exactly 6 successes)=   16C61463410

Page No 33.29:

Question 21:

In a binomial distribution, the probability of getting success is 1/4 and standard deviation is 3. Then, its mean is
(a) 6
(b) 8
(c) 12
(d) 10

Answer:

(c) 12

p =14 q= 1-14=34Standard deviation =3  Variance = 32=9Or  npq = 9 np =9q= 363=12Therefore, the mean of the given binomial distribution is 12.

Page No 33.29:

Question 22:

A coin is tossed 4 times. The probability that at least one head turns up is
(a) 116

(b) 216

(c) 1416

(d) 1516

Answer:

(d) 1516

Let X denote the number of heads obtained in four tosses of a coin .
Then X follows a binomial distribution with
n =4 and p =q =12Distribution is given by P(X=r)=Cr412r124-r P(X=r)=C04120124-0 P(atleast one head turns up)= P(X1) =  1-P(X=0) = 1-124=1516

Page No 33.29:

Question 23:

For a binomial variate X, if n = 3 and P (X = 1) = 8 P (X = 3), then p =
(a) 4/5
(b) 1/5
(c) 1/3
(d) 2/3

Answer:

n =3

P(X=1) =8 P(X=3)           (Given)The distribution is given by P(X=r)=Cr3prq3-rP(X=1)=C13p1q2 and P(X=3)=C33p3q03pq2 = 8p3 8p2  = 3q2  8p2= 3(1-p)2  8p2= 3-6p+3p25p2+6p-3=0p=-6±9610

Hence , it does not match any of the answer choices.

Page No 33.29:

Question 24:

A coin is tossed n times. The probability of getting at least once is greater than 0.8. Then, the least value of n, is
(a) 2
(b) 3
(c) 4
(d) 5

Answer:

(b) 3

Let X be the number of heads. Then X follows a binomial distribution with

p =12,  q=12Hence, the distribution is given by P(X=r) = Crn12r12n-r, r=0,1,2,3......nP(X1) =1-P(X=0)  = 1-12n 0.8Or  2n10.2 2n5This is possible only when n 3.So, the least value of n must be 3.

Page No 33.29:

Question 25:

The probability of selecting a male or a female is same. If the probability that in an office of n persons (n − 1) males being selected is 3210, the value of n is
(a) 5
(b) 3
(c) 10
(d) 12

Answer:

(d) 12

Let X be the number of males. 
p = q = 12      (given)P(X=n-1) =Cn-1npn-1q1  =  3210n12n  = 3210 n12n = 322212   n12n= 121212By comparing the two sides, we get n =12

Page No 33.29:

Question 26:

Mark the correct alternative in the following question:

A box contains 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?

a 9105                                  b 129104                                  c 129105                                  d 9105+129104

Answer:

We have,p=probability of getting a defective pen in a trial=10100=110 andq=probability of getting a good pen in a trial=1-p=1-110=910Let X denote a success of getting a defective pen. Then,X follows a binomial distribution with parameters n=5 and p=110 PX=r=Cr5prq5-r=Cr5110r9105-r=Cr5110595-r=Cr595-r105, where r=0,1,2,3,4,5Now,Required probability=PX1=PX=0+PX=1=C0595105+C1594105=C0595105+C1594105=9105+5×94105=9105+942×104=9105+129104

Hence, the correct alternative is option (d).

Page No 33.29:

Question 27:

Mark the correct alternative in the following question:

Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If PX=rPX=n-r is independent of n and r, then p equals

a 12                                               b 13                                               c 15                                               d 17

Answer:

As, X follows the binomial distribution with parameters n and p, where 0<p<1So, PX=r=Crnprqn-r, where r=0,1,2,3,...,nNow,PX=rPX=n-r=Crnprqn-rCn-rnpn-rqr=p2r-nqn-2rAs, PX=rPX=n-r is independent of n and rSo, p=q                  Since, PX=rPX=n-r=p2r-npn-2r=p0=1This is only possible if p=12 p=12

Hence, the correct alternative is option (a).



Page No 33.30:

Question 28:

Mark the correct alternative in the following question:

The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is

a C450.740.3                                b C150.70.34                                c C450.70.34                                d 0.740.3

Answer:

We have,q=probability that a person is not a swimmer=0.3 andp=probability that a person is a swimmer=1-q=1-0.3=0.7Let X denote a success that a person selected is a swimmer. Then,X follows the binomial distribution with parameters n=5 and p=0.7 PX=r=Cr5prq5-r=Cr50.7r0.35-rNow,The required probability=PX=4=C450.740.35-4=C450.740.3

Hence, the correct alternative is option (a).

Page No 33.30:

Question 29:

Mark the correct alternative in the following question:

Which one is not a requirement of a binomial dstribution?

(a) There are 2 outcomes for each trial
(b) There is a fixed number of trials
(c) The outcomes must be dependent on each other
(d) The probability of success must be the same for all the trials.

Answer:

Since, the trials of the binomial distribution are independent

So, the oucomes should not be dependent on each other

Hence, the correct alternative is option (c).

Page No 33.30:

Question 30:

Mark the correct alternative in the following question:

The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is

a 764                                   b 7128                                   c 451024                                   d 741

Answer:

We have,p=probabiltiy of guessing the answer of a true false correctly=12 andq=probabiltiy of guessing the answer of a true false incorrectly=1-p=1-12=12Let X denote a success of guessing the answer correctly. Then,X follows the binomial distribution with parameters n=10 and p=12 PX=r=Cr10prq10-r=Cr1012r1210-r=Cr101210=Cr10210Now,Required probability=PX8=PX=8+PX=9+PX=10=C810210+C910210+C1010210=45+10+1210=561024=7128

Hence, the correct alternative is option (b).



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