NCERT Solutions for Exercise 13.5 Class 12 Maths Chapter 13 - Probability

Question:1(ii) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of

at least 5 successes?

Answer:

X be a number of success of getting an odd number.

$P(probability\, of\, getting\, odd\, number)=\frac{3}{6}=\frac{1}{2}$

$\therefore \, \, \, \, q=1-P=1-\frac{1}{2}=\frac{1}{2}$

X has a binomial distribution.

$\therefore\ P(X=x)=^nC_n - x.q^{n-x}.p^x$

$P(X=x)=^6C_6-x.$$(\frac{1}{2})^{6-x} . (\frac{1}{2})^{x}$

$P(X=x)=^6C_6-x.$$(\frac{1}{2})^{6}$

$P(At \, \, least \, \, 5\, \, \, success) =P(x\geq 5)$

$=P(X=5)+P(X=6)$

$= ^6C_5 .(\frac{1}{2})^6+^6C_6 .(\frac{1}{2})^6$

$= 6 .(\frac{1}{64})$$+ (\frac{1}{64})$

$= \frac{7}{64}$

Question:1(iii) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of

at most 5 successes?

Answer:

X be a number of success of getting an odd number.

$P(probability\, of\, getting\, odd\, number)=\frac{3}{6}=\frac{1}{2}$

$\therefore \, \, \, \, q=1-P=1-\frac{1}{2}=\frac{1}{2}$

X has a binomial distribution.

$\therefore\ P(X=x)=^nC_n - x.q^{n-x}.p^x$

$P(X=x)=^6C_6-x.$$(\frac{1}{2})^{6-x} . (\frac{1}{2})^{x}$

$P(X=x)=^6C_6-x.$$(\frac{1}{2})^{6}$

$P(atmost\, \, 5\, \, \, success) =P(x\leq 5)$

$=1-P(X> 5)$

$=1-P(X= 5)$

$= 1-^6C_6 .(\frac{1}{2})^6$

$= 1- (\frac{1}{64})$

$= \frac{63}{64}$

Question:2 A pair of dice is thrown $4$ times. If getting a doublet is considered a success, find the probability of two successes

Answer:

A pair of dice is thrown $4$ times.X be getting a doublet.

Probability of getting doublet in a throw of pair of dice :

$P=\frac{6}{36}=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

X has a binomial distribution,n=4

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^4C_x.$$(\frac{5}{6})^{4-x} . (\frac{1}{6})^{x}$

$P(X=x)=^4C_x.$$\frac{5^{4-x}}{6^4}$

Put x = 2

$P(X=2)=^4C_2.$ $\frac{5^{4-2}}{6^4}$

$P(X=2)=6\times \frac{25}{1296}$

$P(X=2)= \frac{25}{216}$

Question:3 There are $5^{o}/_{o}$ defective items in a large bulk of items. What is the probability that a sample of $10$ items will include not more than one defective item?

Answer:

There are $5^{o}/_{o}$ defective items in a large bulk of items.

X denotes the number of defective items in a sample of 10.

$\Rightarrow \, \, P=\frac{5}{100}=\frac{1}{2}$

$\Rightarrow \, \, q=1-\frac{1}{20}=\frac{19}{20}$

X has a binomial distribution, n=10.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^1^0C_x.$$(\frac{19}{20})^{10-x} . (\frac{1}{20})^{x}$

$P(not\, more\, than\, one\, defective item)=p(X\leq 1)$

$=P(X=0)+P(X=1)$

$=^{10}C_0(\frac{19}{20})^{10} . (\frac{1}{20})^{0}+^{10}C_1(\frac{19}{20})^{9} . (\frac{1}{20})^{1}$

$=(\frac{19}{20})^{9} . (\frac{29}{20})^{1}$

Question:4(i) Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards. What is the probability that

all the five cards are spades?

Answer:

Let X represent a number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards.

We have 13 spades.

$P=\frac{13}{52}=\frac{1}{4}$

$q=1-P=1-\frac{1}{4}=\frac{3}{4}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(\frac{3}{4})^{5-x} . (\frac{1}{4})^{x}$

Put X=5 ,

$P(X=5)=^5C_5.$$(\frac{3}{4})^{0} . (\frac{1}{4})^{5}$

$=1\times \frac{1}{1024}$

$= \frac{1}{1024}$

Question:4(ii) Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

only 3 cards are spades?

Answer:

Let X represent a number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards.

We have 13 spades.

$P=\frac{13}{52}=\frac{1}{4}$

$q=1-P=1-\frac{1}{4}=\frac{3}{4}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(\frac{3}{4})^{5-x} . (\frac{1}{4})^{x}$

Put X=3 ,

$P(X=3)=^5C_3.$$(\frac{3}{4})^{2} . (\frac{1}{4})^{3}$

$=10\times \frac{9}{16}\times \frac{1}{64}$

$=\frac{45}{512}$

Question:4(iii) Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

none is a spade?

Answer:

Let X represent number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards.

We have 13 spades .

$P=\frac{13}{52}=\frac{1}{4}$

$q=1-P=1-\frac{1}{4}=\frac{3}{4}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(\frac{3}{4})^{5-x} . (\frac{1}{4})^{x}$

Put X=0 ,

$P(X=0)=^5C_0.$$(\frac{3}{4})^{5} . (\frac{1}{4})^{0}$

$=1\times \frac{243}{1024}$

$= \frac{243}{1024}$

Question:5(i) 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

none will fuse after $150$ days of use.

Answer:

Let X represent number of bulb that will fuse after $150$ days of use .Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(0.95)^{5-x} . (0.05)^{x}$

Put X=0 ,

$P(X=0)=^5C_0.$$(0.95)^{5} . (0.05)^{0}$

$=(0.95)^{5}$

Question:5(ii) 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

not more than one will fuse after $150$ days of use.

Answer:

Let X represent a number of the bulb that will fuse after $150$ days of use. Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(0.95)^{5-x} . (0.05)^{x}$

Put $X\leq 1$ ,

$P(X\leq 1)=P(X=0)+P(X=1)$

$=^5C_0.$$(0.95)^{5} . (0.05)^{0}+^5C_1 (0.95)^{4} . (0.05)^{1}$

$=(0.95)^{5}+ (0.25)(0.95)^4$

$=(0.95)^{4}(0.95+ 0.25)$

$=(0.95)^{4}\times 1.2$

Question:5(iii) 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

more than one will fuse after$150$ days of use.

Answer:

Let X represent number of bulb that will fuse after $150$ days of use .Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(0.95)^{5-x} . (0.05)^{x}$

Put $X> 1$ ,

$P(X> 1)=1-(P(X=0)+P(X=1))$

$=1-(^5C_0.$$(0.95)^{5} . (0.05)^{0}+^5C_1 (0.95)^{4} . (0.05)^{1})$

$=1-((0.95)^{5}+ (0.25)(0.95)^4)$

$=1-((0.95)^{4}(0.95+ 0.25))$

$=1-(0.95)^{4}\times 1.2$

Question:5(iv) 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

at least one will fuse after $150$ days of use.

Answer:

Let X represent number of bulb that will fuse after $150$ days of use .Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(0.95)^{5-x} . (0.05)^{x}$

Put $X\geq 1$ ,

$P(X\geq 1)=1-P(X< 1)$

$P(X\geq 1)=1-P(X=0)$

$=1-^5C_0.$$(0.95)^{5} . (0.05)^{0}$

$=1-(0.95)^{5}$

Question:6 A bag consists of $10$ balls each marked with one of the digits $0$ 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 denote a number of balls marked with digit 0 among 4 balls drawn.

Balls are drawn with replacement.

X has a binomial distribution,n=4.

$P=\frac{1}{10}$

$q=1-P=1-\frac{1}{10}=\frac{9}{10}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^4C_x.$$(\frac{9}{10})^{4-x} . (\frac{1}{10})^{x}$

Put X = 0,

$P(X=0)=^4C_0.$$(\frac{9}{10})^{4} . (\frac{1}{10})^{0}$

$= 1.(\frac{9}{10})^{4}$

$= (\frac{9}{10})^4$

Question:7 In an examination,$20$ questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false'. Find the probability that he answers at least 12 questions correctly.

Answer:

Let X represent the number of correctly answered questions out of 20 questions.

The coin falls heads, he answers 'true'; if it falls tails, he answers 'false'.

$P=\frac{1}{2}$

$q=1-P=1-\frac{1}{2}=\frac{1}{2}$

X has a binomial distribution,n=20

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^2^0C_x.$$(\frac{1}{2})^{20-x} . (\frac{1}{2})^{x}$

$P(X=x)=^2^0C_x.$ $(\frac{1}{2})^{20}$

$P(at\, \, least\, 12\, \,questions \, \, answered\, \, correctly)=P(X\geq 12)$

$=P(X=12)+P(X=13)..................+P(X=20)$

$=^{20}C_1_2 (\frac{1}{2})^{20}+^{20}C_1_3(\frac{1}{2})^{20}+..........^{20}C_2_0 (\frac{1}{2})^{20}$

$=(\frac{1}{2})^{20}(^{20}C_1_2 +^{20}C_1_3+..........^{20}C_2_0 )$

Question:8 Suppose X has a binomial distribution $B\left [ 6,\frac{1}{2} \right ].$ Show that $X=3$ is the most likely outcome.

(Hint : $P(X=3)$ is the maximum among all of $P(x_{i})$ ,$x_{i}=0,1,2,3,4,5,6$ )

Answer:

X is a random variable whose binomial distribution is $B\left [ 6,\frac{1}{2} \right ].$

Here , n=6 and $P=\frac{1}{2}$.

$\therefore \, \,q=1-P= 1-\frac{1}{2}= \frac{1}{2}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$=^{6}C_x .(\frac{1}{2})^{6-x}(\frac{1}{2})^x$

$=^{6}C_x (\frac{1}{2})^6$

$P(X=x)$ is maximum if $^{6}C_x$ is maximum.

$^{6}C_0 =^{6}C_6 =\frac{6!}{0!.6!}=1$

$^{6}C_1 =^{6}C_5 =\frac{6!}{1!.5!}=6$

$^{6}C_2 =^{6}C_4 =\frac{6!}{2!.4!}=15$

$^{6}C_3 =\frac{6!}{3!.3!}=20$

$^{6}C_3$ is maximum so for x=3 ,$P(X=3)$ is maximum.

Question:9 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?

Answer:

Let X represent number of correct answers by guessing in set of 5 multiple choice questions.

Probability of getting a correct answer :

$P=\frac{1}{3}$

$\therefore q=1-P=1-\frac{1}{3}=\frac{2}{3}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(\frac{2}{3})^{5-x} . (\frac{1}{3})^{x}$

$P(guessing \, \, more\, \, than \, 4\, correct\, answer)=P(X\geq 1)$

$=P(X=4)+P(X=5)$

$=^5C_4.$$(\frac{2}{3})^{1} . (\frac{1}{3})^{4}+^5C_5(\frac{2}{3})^{0} . (\frac{1}{3})^{5}$

$=\frac{10}{243}+\frac{1}{243}$

$=\frac{11}{243}$

Question:10(a) A person buys a lottery ticket in $50$ lotteries, in each of which his chance of winning a prize is $\frac{1}{100}.$ What is the probability that he will win a prize

at least once

Answer:

Let X represent number of winning prizes in 50 lotteries .

$P=\frac{1}{100}$

$q=1-P=1-\frac{1}{100}=\frac{99}{100}$

X has a binomial distribution,n=50.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^50C_x.$$(\frac{1}{100})^{50-x} . (\frac{99}{100})^{x}$

$P(winning \:at\: least\: once)=P(X\geq 1)$

$=1-P(X< 1)$

$=1-P(X= 0)$

$=1- ^{50}C_0 (\frac{99}{100})^{50}$

$=1- 1. (\frac{99}{100})^{50}$

$=1- (\frac{99}{100})^{50}$

Question:10(b) A person buys a lottery ticket in $50$ lotteries, in each of which his chance of winning a prize is $\frac{1}{100}$. What is the probability that he will win a prize

exactly once

Answer:

Let X represent number of winning prizes in 50 lotteries .

$P=\frac{1}{100}$

$q=1-P=1-\frac{1}{100}=\frac{99}{100}$

X has a binomial distribution,n=50.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^50C_x.$$(\frac{1}{100})^{50-x} . (\frac{99}{100})^{x}$

$P(winning\, exactly\, once)=P(X= 1)$

$=^{50}C_1 (\frac{99}{100})^{49}.\frac{1}{100}$

$= 50.(\frac{99}{100})^{49}\frac{1}{100}$

$=\frac{1}{2} (\frac{99}{100})^{50}$

Question:10(c) A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is $\frac{1}{100}$ . What is the probability that he will win a prize

at least twice?

Answer:

Let X represent number of winning prizes in 50 lotteries.

$P=\frac{1}{100}$

$q=1-P=1-\frac{1}{100}=\frac{99}{100}$

X has a binomial distribution,n=50.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^50C_x.$$(\frac{1}{100})^{50-x} . (\frac{99}{100})^{x}$

$P(winning \, \, at \, \, least \, \, twice)=P(X\geq 2)$

$=1-P(X< 2)$

$=1-P(X\leq 1)$

$=1-(P(X=0)+P(X=1))$

$=1- (\frac{99}{100})^{50}-\frac{1}{2}(\frac{99}{100})^{49}$

$=1- (\frac{99}{100})^{49}(\frac{1}{2}+\frac{99}{100})$

$=1- (\frac{99}{100})^{49}\frac{149}{100}$

Question:11 Find the probability of getting $5$ exactly twice in $7$ throws of a die.

Answer:

Let X represent number of times getting 5 in 7 throws of a die.

Probability of getting 5 in single throw of die=P

$P=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

X has a binomial distribution,n=7

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^7C_x.$$(\frac{5}{6})^{7-x} . (\frac{1}{6})^{x}$

$P(getting\, \, exactly\, \, twice)=P(X= 2)$

$=^{7}C_2 (\frac{5}{6})^{5}\frac{1}{6}^2$

$=21 (\frac{5}{6})^{5}\frac{1}{36}$

$= (\frac{5}{6})^{5}\frac{7}{12}$

Question:12 Find the probability of throwing at most $2$ sixes in $6$ throws of a single die.

Answer:

Let X represent number of times getting 2 six in 6 throws of a die.

Probability of getting 6 in single throw of die=P

$P=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

X has a binomial distribution,n=6

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^6C_x.$$(\frac{5}{6})^{6-x} . (\frac{1}{6})^{x}$

$P(getting\, \, atmost\, \, two\, six)=P(X\leq 2)$

$=P(X=0)+P(X=1)+P(X=2)$

$=^{6}C_0 (\frac{5}{6})^{6}\frac{1}{6}^0+^{6}C_1 (\frac{5}{6})^{5}\frac{1}{6}^1+^{6}C_2 (\frac{5}{6})^{4}\frac{1}{6}^2$

$=1.(\frac{5}{6})^{6}+6. (\frac{5}{6})^{5}\frac{1}{6}+15 (\frac{5}{6})^{4}\frac{1}{36}$

$=(\frac{5}{6})^{6}+ (\frac{5}{6})^{5}+ (\frac{5}{6})^{4}\frac{5}{12}$

$=(\frac{5}{6})^{4}( \frac{5}{6}^{2}+ \frac{5}{6}+\frac{5}{12})$

$=(\frac{5}{6})^{4}( \frac{25}{36}+ \frac{5}{6}+\frac{5}{12})$

$=(\frac{5}{6})^{4}( \frac{70}{36})$

$=(\frac{5}{6})^{4}( \frac{35}{18})$

Question:13 It is known that $10^{o}/_{o}$ of certain articles manufactured are defective. What is the probability that in a random sample of $12$ such articles,$9$ are defective?

Answer:

Let X represent a number of times selecting defective items out of 12 articles.

Probability of getting a defective item =P

$P=10\%=\frac{10}{100}=\frac{1}{10}$

$q=1-P=1-\frac{1}{10}=\frac{9}{10}$

X has a binomial distribution,n=12

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^1^2C_x.$$(\frac{9}{10})^{12-x} . (\frac{1}{10})^{x}$

$P(selectting\, \, 9\,defective \, items)=$

$=^{12}C_9 (\frac{9}{10})^{3}\frac{1}{10}^9$

$=220 (\frac{9^3}{10^3})\frac{1}{10^9}$

$=22 \times \frac{9^3}{10^1^1}$

Question:14 In a box containing $100$ bulbs, $10$ are defective. The probability that out of a sample of $5$ bulbs, none is defective is

(A) $10^{-1}$

(B) $\left ( \frac{1}{5} \right )^{5}$

(C) $\left ( \frac{9}{10} \right )^{5}$

(D) $\frac{9}{10}$

Answer:

Let X represent a number of defective bulbs out of 5 bulbs.

Probability of getting a defective bulb =P

$P=\frac{10}{100}=\frac{1}{10}$

$q=1-P=1-\frac{1}{10}=\frac{9}{10}$

X has a binomial distribution,n=5

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(\frac{9}{10})^{5-x} . (\frac{1}{10})^{x}$

$P(non\,of \,bulb\,is\, defective \,)=P(X=0)$

$=^{5}C_0 (\frac{9}{10})^{5}\frac{1}{10}^0$

$=1.\frac{9}{10}^5$

$=(\frac{9}{10})^5$

The correct answer is C.

Question:15 The probability that a student is not a swimmer is $\frac{1}{5}.$ Then the probability that out of five students, four are swimmers is

In the following, choose the correct answer:

(A) $^{5}C_{4}\left ( \frac{4}{5} \right )^{4}\frac{1}{5}$

(B) $\left ( \frac{4}{5} \right )^{4}\frac{1}{5}$

(C) $^{5}C_{1}\frac{1}{5}\left ( \frac{4}{5} \right )^{4}$

(D) None of these

Answer:

Let X represent number students out of 5 who are swimmers.

Probability of student who are not swimmers =q

$q=\frac{1}{5}$

$P=1-q=1-\frac{1}{5}=\frac{4}{5}$

X has a binomial distribution,n=5

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$$(\frac{1}{5})^{5-x} . (\frac{4}{5})^{x}$

$P(4\, \,students\, \, are\, swimmers)=P(X= 4)$

$=^5C_4(\frac{1}{5})^{1} . (\frac{4}{5})^{4}$

Option A is correct.