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

Exercise 13.5

1. A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 success? (iii) at most 5 successes?

Ans: The given experiment of tossing a die is a Bernoulli trial. 

Let \[X\] be the number of successes occurred in an experiment of getting odd numbers in 6 trials.

Probability of an experiment is ratio of number of favourable outcomes to the total number of outcomes.

Set of odd numbers when a die is rolled is given by,$\left\{ {1,3,5} \right\}$.

Therefore, number of favourable outcomes is 3.

Probability of getting an odd number by performing first trial is, 

$p = \dfrac{3}{6}$

$p = \dfrac{1}{2}$

Probability of failure is calculated by using the formula $q = 1 - p$. 

Probability of not getting an odd number is given by,

$q = 1 - \dfrac{1}{2}$

$q = \dfrac{1}{2}$

Random variable $X$ follows binomial distribution with $p = \dfrac{1}{2}$ and $n = 6$

Probability is calculated using the formula,

$P\left( {X = x} \right) = {}^n{C_x}{q^{n - x}}{p^x}$

Substitute values of $p,q$ and $n$ in the above formula.

$P\left( {X = x} \right) = {}^6{C_x}{\left( {\dfrac{1}{2}} \right)^{6 - x}}{\left( {\dfrac{1}{2}} \right)^x}$

$P\left( {X = x} \right) = {}^6{C_x}{\left( {\dfrac{1}{2}} \right)^{6 - x}}{\left( {\dfrac{1}{2}} \right)^x}$

(i). The probability of 5 successes.

To find probability of 5 successes substitute 5 for $x$ in the probability formula.

$P\left( {X = 5} \right) = {}^6{C_5}{\left( {\dfrac{1}{2}} \right)^{6 - 5}}{\left( {\dfrac{1}{2}} \right)^5}$

$P\left( {X = 5} \right) = {}^6{C_5}\left( {\dfrac{1}{2}} \right){\left( {\dfrac{1}{2}} \right)^5}$

$P\left( {X = 5} \right) = 6 \times \dfrac{1}{{64}}$

$P\left( {X = 5} \right) = \dfrac{3}{{32}}$

Therefore, probability of getting 5 successes is $\dfrac{3}{{32}}$.

(ii). Probability of at least 5 successes.

To find probability of at least 5 successes find the probability $P\left( {X \geqslant 5} \right)$. Add probabilities $P\left( {X = 5} \right)$and $P\left( {X = 6} \right)$.

$P\left( {X \geqslant 5} \right) = P\left( {X = 5} \right) + P\left( {X = 6} \right)$

$P\left( {X \geqslant 5} \right) = {}^6{C_5}\left( {\dfrac{1}{2}} \right){\left( {\dfrac{1}{2}} \right)^5} + {}^6{C_6}{\left( {\dfrac{1}{2}} \right)^6}$

$P\left( {X \geqslant 5} \right) = \dfrac{6}{{64}} + \dfrac{1}{{64}}$

$P\left( {X \geqslant 5} \right) = \dfrac{7}{{64}}$

Therefore, probability of at least 5 successes is $\dfrac{7}{{64}}$.

(iii). Probability of at most 5 successes

To find probability of at most 5 successes. Find the probability $P\left( {X \leqslant 5} \right)$. For that Subtract probability $P\left( {X > 5} \right)$ from 1.

$P\left( {X \leqslant 5} \right) = 1 - P\left( {X > 5} \right)$

$P\left( {X \leqslant 5} \right) = 1 - P\left( {X = 6} \right)$

$P\left( {X \leqslant 5} \right) = 1 - {}^6{C_6}{\left( {\dfrac{1}{2}} \right)^6}$ 

On evaluating further,

$P\left( {X \leqslant 5} \right) = 1 - \dfrac{1}{{64}}$

$P\left( {X \leqslant 5} \right) = \dfrac{{64 - 1}}{{64}}$

$P\left( {X \leqslant 5} \right) = \dfrac{{63}}{{64}}$

Therefore, probability of at most 5 successes is $\dfrac{{63}}{{64}}$.

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

Ans:The given experiment of tossing two dice is a Bernoulli trial. 

Let $X$ be the number of successes occurred in an experiment of getting a doublet in 4 trials.

The probability of an experiment is supposed to be a ratio of number of favourable outcomes to the total number of outcomes

Set of doublets when two dice are rolled is given by, $\left\{ {\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {4,4} \right),\left( {5,5} \right),\left( {6,6} \right)} \right\}$.

Therefore, number of favourable outcomes is 6.

Probability of getting a doublet by performing first trial is, 

$p = \dfrac{6}{{36}}$

$p = \dfrac{1}{6}$

Probability of failure is calculated by using the formula $q = 1 - p$. 

Probability of not getting an odd number is given by,

$q = 1 - \dfrac{1}{6}$

$q = \dfrac{5}{6}$

Random variable $X$ follows binomial distribution with $p = \dfrac{1}{6}$ and $n = 4$

Probability is calculated using the formula,

$P\left( {X = x} \right) = {}^n{C_x}{q^{n - x}}{p^x}$

Substitute values of $p,q$ and $n$ in the above formula.

$P\left( {X = x} \right) = {}^4{C_x}{\left( {\dfrac{5}{6}} \right)^{4 - x}}{\left( {\dfrac{1}{6}} \right)^x}$

$P\left( {X = x} \right) = {}^4{C_x}{\left( {\dfrac{5}{6}} \right)^{4 - x}}{\left( {\dfrac{1}{6}} \right)^x}$

To find probability of 2 successes substitute 2 for $x$ in the probability formula.

$P\left( {X = 2} \right) = {}^4{C_2}{\left( {\dfrac{5}{6}} \right)^{4 - 2}}{\left( {\dfrac{1}{6}} \right)^2}$

$P\left( {X = 2} \right) = {}^4{C_2}{\left( {\dfrac{5}{6}} \right)^2}{\left( {\dfrac{1}{6}} \right)^2}$

$P\left( {X = 2} \right) = 6 \times \dfrac{{25}}{{1296}}$

$P\left( {X = 2} \right) = \dfrac{{25}}{{216}}$

Therefore, probability of getting 2 successes is $\dfrac{{25}}{{216}}$.

3. There are $5\% $ 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?

Ans: It is given that in a large bulk of items $5\% $ are defective. As, the items are drawn with replacement, the trials are found to be Bernoulli trials. 

Consider $X$ to be a variable that represents the number of defective items in a sample of 10 items drawn successively.

The probability of an experiment is supposed to be a ratio of number of favourable outcomes to the total number of outcomes

Thus,

$p = \dfrac{5}{{100}}$, that is, $p = \dfrac{1}{{20}}$

Now, $q = 1 - p$

$q = 1 - \dfrac{1}{{20}}$

$q = \dfrac{{19}}{{20}}$

It can be seen that the variable $X$ is found to have a binomial distribution with $n = 10$ and  $p = \dfrac{1}{{20}}$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Where, $x = 0,1,2,.......,n$.

Substitute 10 for $n$, $\dfrac{1}{{20}}$ for $p$ and $\dfrac{{19}}{{20}}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^{10}}{C_x}{\left( {\dfrac{{19}}{{20}}} \right)^{10 - x}}{\left( {\dfrac{1}{{20}}} \right)^x}$ 

The probability that a sample of 10 items will include not more than one defective item is calculated as follows,

$P\left( {{\text{not more than 1 defective item}}} \right) = P\left( {X \leqslant 1} \right)$

$P\left( {X \leqslant 1} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right)$

Substitute the values of $x$ as 0 and 1 in the above formula,

$P\left( {X \leqslant 1} \right){ = ^{10}}{C_0}{\left( {\dfrac{{19}}{{20}}} \right)^{10}}{\left( {\dfrac{1}{{20}}} \right)^0}{ + ^{10}}{C_1}{\left( {\dfrac{{19}}{{20}}} \right)^9}{\left( {\dfrac{1}{{20}}} \right)^1}$

The value of $^10{C_0} = 1$ and $^10{C_1} = 10$.

$P\left( {X \leqslant 1} \right) = \left( 1 \right){\left( {\dfrac{{19}}{{20}}} \right)^{10}}\left( 1 \right) + 10{\left( {\dfrac{{19}}{{20}}} \right)^9}\left( {\dfrac{1}{{20}}} \right)$

\[P\left( {X \leqslant 1} \right) = {\left( {\dfrac{{19}}{{20}}} \right)^{10}} + 10{\left( {\dfrac{{19}}{{20}}} \right)^9}\left( {\dfrac{1}{{20}}} \right)\]

On taking out the common term,

\[P\left( {X \leqslant 1} \right) = {\left( {\dfrac{{19}}{{20}}} \right)^{10}} + 10{\left( {\dfrac{{19}}{{20}}} \right)^9}\left( {\dfrac{1}{{20}}} \right)\]

\[P\left( {X \leqslant 1} \right) = {\left( {\dfrac{{19}}{{20}}} \right)^9}\left[ {\left( {\dfrac{{19}}{{20}}} \right) + \left( {\dfrac{{10}}{{20}}} \right)} \right]\]

\[P\left( {X \leqslant 1} \right) = {\left( {\dfrac{{19}}{{20}}} \right)^9}\left[ {\dfrac{{29}}{{20}}} \right]\]

\[P\left( {X \leqslant 1} \right) = \left( {\dfrac{{29}}{{20}}} \right){\left( {\dfrac{{19}}{{20}}} \right)^9}\]

Therefore, the probability that a sample of 10 items that will include not more than one defective item is \[\left( {\dfrac{{29}}{{20}}} \right){\left( {\dfrac{{19}}{{20}}} \right)^9}\].

4. Five cards are drawn successively with replacement from a well-shuffled deck 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?

Ans: It is given that five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. As, the cards are drawn with replacement, the trials are found to be Bernoulli trials.

Consider a variable $X$ that represents the number of spade cards among the five cards drawn.

Out of 52 cards, it is known that 13 spade cards.

The probability of an experiment is supposed to be a ratio of number of favourable outcomes to the total number of outcomes

Thus,

$p = \dfrac{{13}}{{52}}$

$p = \dfrac{1}{4}$

Now, $q = 1 - p$

$q = 1 - \dfrac{1}{4}$

$q = \dfrac{3}{4}$

It can be seen that the variable $X$ is found to have a binomial distribution with $n = 5$  and  $p = \dfrac{1}{4}$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Where, $x = 0,1,2,.......,n$.

Substitute 5 for $n$, $\dfrac{1}{4}$ for $p$ and $\dfrac{3}{4}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^5}{C_x}{\left( {\dfrac{3}{4}} \right)^{5 - x}}{\left( {\dfrac{1}{4}} \right)^x}$

Now, consider the conditions.

(i) Probability that all five cards are spades.

$P\left( {{\text{all five cards are spades}}} \right) = P\left( {X = 5} \right)$

$P\left( {X = 5} \right){ = ^5}{C_5}{\left( {\dfrac{3}{4}} \right)^{5 - 5}}{\left( {\dfrac{1}{4}} \right)^5}$

$P\left( {X = 5} \right) = \left( 1 \right){\left( {\dfrac{3}{4}} \right)^0}{\left( {\dfrac{1}{4}} \right)^5}$

$P\left( {X = 5} \right) = \left( 1 \right)\left( 1 \right){\left( {\dfrac{1}{4}} \right)^5}$

$P\left( {X = 5} \right) = \dfrac{1}{{1024}}$

Therefore, the probability that all five cards are spades is $\dfrac{1}{{1024}}$.

(ii) Probability that only 3 cards are spades.

$P\left( {{\text{only 3 cards are spades}}} \right) = P\left( {X = 3} \right)$

$P\left( {X = 3} \right){ = ^5}{C_3}{\left( {\dfrac{3}{4}} \right)^{5 - 3}}{\left(     {\dfrac{1}{4}} \right)^3}$

The formula for $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

$P\left( {X = 3} \right) = \dfrac{{5!}}{{3!2!}}{\left( {\dfrac{3}{4}} \right)^2}{\left( {\dfrac{1}{4}} \right)^3}$

$P\left( {X = 3} \right) = 10\left( {\dfrac{9}{{16}}} \right)\left( {\dfrac{1}{{64}}} \right)$

On evaluating further,

$P\left( {X = 3} \right) = \dfrac{{45}}{{512}}$

Therefore, the probability that only 3 cards are spades is found to be $\dfrac{{45}}{{512}}$.

(iii) Probability that none is a spade.

$P\left( {{\text{none is a spade}}} \right) = P\left( {X = 0} \right)$

$P\left( {X = 0} \right){ = ^5}{C_0}{\left( {\dfrac{3}{4}} \right)^{5 - 0}}{\left( {\dfrac{1}{4}} \right)^0}$

The formula for $^5{C_0} = 1$.

$P\left( {X = 0} \right) = \left( 1 \right){\left( {\dfrac{3}{4}} \right)^5}\left( 1 \right)$

$P\left( {X = 0} \right) = {\left( {\dfrac{3}{4}} \right)^5}$

On evaluating further,

$P\left( {X = 0} \right) = \left( {\dfrac{{243}}{{1024}}} \right)$

Therefore, the probability that none is a spade is $\dfrac{{243}}{{1024}}$.

5. The probability that a bulb produced by a factory will fuse after 150 days of use is $0.05$. What is the probability that out of 5 such bulbs 

1. none 

2. not more than one 

3. more than one 

4. at least one 

Will fuse after 150 days of use.

Ans: It is given that the probability that a bulb produced by a factory will fuse after 150 days of use is $0.05$. The trials are found to be Bernoulli trials as replacement occurs in the event.

Consider $X$ to be a variable that represents the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials.

Now, $p = 0.05$.

As, $q = 1 - p$

$q = 1 - 0.05$

$q = 0.95$

It can be seen that the variable $X$ is found to have a binomial distribution with $n = 5$  and  $p = 0.05$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Where, $x = 0,1,2,.......,n$.

Substitute 5 for $n$, $0.05$ for $p$ and $0.95$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^5}{C_x}{\left( {0.95} \right)^{5 - x}}{\left( {0.05} \right)^x}$

Now, consider the conditions,

(i). Probability that none will fuse after 150 days of use.

$P\left( {{\text{none will fuse after the 150 days of use}}} \right) = P\left( {X = 0} \right)$

$P\left( {X = 0} \right){ = ^5}{C_0}{\left( {0.95} \right)^5}{\left( {0.05} \right)^0}$

The value of $^5{C_0} = 1$.

$P\left( {X = 0} \right) = \left( 1 \right){\left( {0.95} \right)^5}\left( 1 \right)$

$P\left( {X = 0} \right) = {\left( {0.95} \right)^5}$

Therefore, the probability that out of 5 such bulb none will fuse after 150 days of use is ${\left( {0.95} \right)^5}$.

(ii). Probability that not more than one bulb will fuse after 150 days of use.

$P\left( {{\text{not more than one bulb will fuse after the 150 days of use}}} \right) = P\left( {X \leqslant 1} \right)$

$P\left( {X \leqslant 1} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right)$

$P\left( {X \leqslant 1} \right){ = ^5}{C_0}{\left( {0.95} \right)^5}{\left( {0.05} \right)^0}{ + ^5}{C_1}{\left( {0.95} \right)^4}{\left( {0.05} \right)^1}$

The value of $^5{C_0} = 1$ and $^5{C_1} = 5$.

$P\left( {X \leqslant 1} \right) = 1{\left( {0.95} \right)^5}\left( 1 \right) + 5{\left( {0.95} \right)^4}\left( {0.05} \right)$

$P\left( {X \leqslant 1} \right) = {\left( {0.95} \right)^5} + 5{\left( {0.95} \right)^4}\left( {0.05} \right)$

$P\left( {X \leqslant 1} \right) = {\left( {0.95} \right)^5} + \left( {0.25} \right){\left( {0.95} \right)^4}$

Now, take out the common factor,

$P\left( {X \leqslant 1} \right) = {\left( {0.95} \right)^4}\left[ {0.95 + 0.25} \right]$

$P\left( {X \leqslant 1} \right) = {\left( {0.95} \right)^4}\left( {1.2} \right)$

Therefore, the probability that out of 5 such bulb not more than one will fuse after 150 days of use is ${\left( {0.95} \right)^4}\left( {1.2} \right)$.

(iii). Probability that more than one bulb will fuse after 150 days of use.

$P\left( {{\text{more than 1 will fuse after the 150 days of use}}} \right) = P(X>1) $

$P(X>1) = 1 - P\left( {X \leqslant 1} \right)$

That is,

$ =1-[(0.95)^4 \times 1.2] $

Therefore, the probability that out of 5 such bulb more than one will fuse after 150 days of use is $1 - {\left( {0.95} \right)^4}\left( {1.2} \right)$.

(iv). Probability that at least one bulb will fuse after 150 days of use. 

$P\left( {{\text{at least one}}} \right) = P\left( {X \geqslant 1} \right)$

$P\left( {X \geqslant 1} \right) = 1 - P\left( {X < 1} \right)$

That is,

$P\left( {X \geqslant 1} \right) = 1 - P\left( {X < 1} \right)$

$P\left( {X \geqslant 1} \right) = 1{ - ^5}{C_0}{\left( {0.95} \right)^5} \times {\left( {0.05} \right)^0}$

On further solving,

$P\left( {X \geqslant 1} \right) = 1{ - ^5}{C_0}{\left( {0.95} \right)^5} \times {\left( {0.05} \right)^0}$

The value of $^5{C_0} = 1$.

$P\left( {X \geqslant 1} \right) = 1 - \left( 1 \right){\left( {0.95} \right)^5} \times \left( 1 \right)$

$P\left( {X \geqslant 1} \right) = 1 - {\left( {0.95} \right)^5}$

Therefore, the probability that out of 5 such bulb at least one will fuse after 150 days of use is$1 - {\left( {0.95} \right)^5}$.

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?

Ans: Consider $X$ be a variable that represents the number of balls marked with the digit 0 among the 4 balls drawn.

As the drawing of balls is an event which involves replacement, the trials are found to be Bernoulli trials.

It can be seen that the variable $X$ is found to have a binomial distribution with $n = 4$ and $p = \dfrac{1}{{10}}$.

Now, it is known that, $q = 1 - p$.

$q = 1 - \dfrac{1}{{10}}$

$q = \dfrac{9}{{10}}$

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Where, $x = 0,1,2,.......,n$.

Substitute 4 for $n$, $\dfrac{1}{{10}}$ for $p$ and $\dfrac{9}{{10}}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^4}{C_x}{\left( {\dfrac{9}{{10}}} \right)^{4 - x}}{\left( {\dfrac{1}{{10}}} \right)^x}$

Now, the probability that none is marked with the digit 0 is calculated as,

$P\left( {{\text{none marked with 0}}} \right) = P\left( {X = 0} \right)$

$P\left( {X = 0} \right){ = ^4}{C_0}{\left( {\dfrac{9}{{10}}} \right)^{4 - 0}}{\left( {\dfrac{1}{{10}}} \right)^0}$

The value of $^4{C_0} = 1$.

$P\left( {X = 0} \right){ = ^4}{C_0}{\left( {\dfrac{9}{{10}}} \right)^{4 - 0}}{\left( {\dfrac{1}{{10}}} \right)^0}$

$P\left( {X = 0} \right) = {\left( {\dfrac{9}{{10}}} \right)^4}{\left( {\dfrac{1}{{10}}} \right)^0}$

$P\left( {X = 0} \right) = {\left( {\dfrac{9}{{10}}} \right)^4}$

Therefore, the probability that none is marked with the digit 0 is found to be ${\left( {\dfrac{9}{{10}}} \right)^4}$.

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 tail, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

Ans: Say, X represents the number of correctly answered questions out of 20 questions.

As the coins are tossed repeatedly, and since, heads represent the true answer and tail represents the false answer. The correctly answered questions are Bernoulli trials.

The probability of an experiment is supposed to be a ratio of number of favourable outcomes to the total number of outcomes.

$p = \dfrac{1}{2}$

Now, it is known that, $q = 1 - p$.

$q = 1 - \dfrac{1}{2}$

$q = \dfrac{1}{2}$

It can be clearly seen that the variable $X$ is found to have a binomial distribution with $n = 20$ and $p = \dfrac{1}{2}$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Where, $x = 0,1,2,.......,n$.

Substitute 20 for $n$, $\dfrac{1}{2}$ for $p$ and $\dfrac{1}{2}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^{20}}{C_x}{\left( {\dfrac{1}{2}} \right)^{20 - x}}{\left( {\dfrac{1}{2}} \right)^x}$

$P\left( {X = x} \right){ = ^{20}}{C_x}{\left( {\dfrac{1}{2}} \right)^{20}}$

Now, the probability that he answers at least 12 questions correctly is calculated as,

$P\left( {{\text{at least 12 questions answered correctly}}} \right) = P\left( {X \geqslant 12} \right)$

$P\left( {X \geqslant 12} \right) = P\left( {X = 12} \right) + P\left( {X = 13} \right) + .... + P\left( {X = 20} \right)$

$P\left( {X \geqslant 12} \right){ = ^{20}}{C_{12}}{\left( {\dfrac{1}{2}} \right)^{20}}{ + ^{20}}{C_{13}}{\left( {\dfrac{1}{2}} \right)^{20}} + ....{ + ^{20}}{C_{20}}{\left( {\dfrac{1}{2}} \right)^{20}}$

$P\left( {X \geqslant 12} \right) = {\left( {\dfrac{1}{2}} \right)^{20}} \cdot \left[ {^{20}{C_{12}}{ + ^{20}}{C_{13}} + ...{ + ^{20}}{C_{20}}} \right]$

Therefore, the probability that he answers at least 12 questions correctly is found to be ${\left( {\dfrac{1}{2}} \right)^{20}} \cdot \left[ {^{20}{C_{12}}{ + ^{20}}{C_{13}} + ...{ + ^{20}}{C_{20}}} \right]$.

8. Suppose $X$ has a binominal distribution $B\left( {6,\dfrac{1}{2}} \right)$.Show that $X = 3$ is the most likely outcome. (Hint: $P\left( {X = 3} \right)$ is the maximum among all $P\left( {{x_i}} \right)$, ${x_i} = 0,1,2,3,4,5,6$)

Ans: It is given that $X$ has a binomial distribution $B\left( {6,\dfrac{1}{2}} \right)$, thus, $n = 6$ and $p = \dfrac{1}{2}$.

Now, it is known that, $q = 1 - p$.

$q = 1 - \dfrac{1}{2}$

$q = \dfrac{1}{2}$

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 6 for $n$,  $\dfrac{1}{2}$ for $p$ and  $\dfrac{1}{2}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^6}{C_x}{\left( {\dfrac{1}{2}} \right)^{6 - x}}{\left( {\dfrac{1}{2}} \right)^x}$

$P\left( {X = x} \right){ = ^6}{C_x}{\left( {\dfrac{1}{2}} \right)^6}$

The value of $P\left( {X = x} \right)$ will be a maximum value, if $^6{C_x}$ will be maximum.

Now, the formula of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Thus, $^6{C_0}{ = ^6}{C_6} = \dfrac{{6!}}{{0!6!}} = 1$

Also, $^6{C_{_1}} = \dfrac{{6!}}{{1!5!}} = 6=^6{C_{_5}}$

$^6{C_2} = \dfrac{{6!}}{{2!4!}} = \dfrac{{6\left( 5 \right)}}{2} = 15=^6{C_{_4}}$

And, 

$^6{C_3} = \dfrac{{6!}}{{3!3!}} = \dfrac{{6 \cdot 5 \cdot 4}}{{3 \cdot 2}} = 20$

It is observed that the value of $^6{C_3}$ is found to be maximum. Therefore, for $x = 3$, the value of $P\left( {X = x} \right)$ is found to be maximum and thus, $X = 3$ is the most likely outcome is proved.

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?

Ans: In multiple choice questions the repeated guessing of the correct answers are Bernoulli trials. Assume that $X$ represents the number of correct answers by guessing in a set of five questions.

The probability of an experiment is supposed to be a ratio of number of favourable outcomes to the total number of outcomes.

Thus, the probability of getting a correct answer is, $p = \dfrac{1}{3}$.

Now,

As, $q = 1 - p$

$q = 1 - \dfrac{1}{3}$

$q = \dfrac{2}{3}$

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 5 for $n$, $\dfrac{1}{3}$ for $p$ and $\dfrac{2}{3}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^5}{C_x}{\left( {\dfrac{2}{3}} \right)^{5 - x}}{\left( {\dfrac{1}{3}} \right)^x}$ 

The probability of guessing more than 4 correct answers is given by,

$P\left( {X \geqslant 4} \right)$

Thus,

$P\left( {X \geqslant 4} \right) = P\left( {X = 4} \right) + P\left( {X = 5} \right)$

$P\left( {X \geqslant 4} \right){ = ^5}{C_4}\left( {\dfrac{2}{3}} \right){\left( {\dfrac{1}{3}} \right)^4}{ + ^5}{C_5}{\left( {\dfrac{1}{3}} \right)^5}$

It is known that, $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

$P\left( {X \geqslant 4} \right) = 5 \cdot \dfrac{2}{3} \cdot \dfrac{1}{{81}} + 1 \cdot \left( {\dfrac{1}{{243}}} \right)$

$P\left( {X \geqslant 4} \right) = \dfrac{{10}}{{243}} + \dfrac{1}{{243}}$

$P\left( {X \geqslant 4} \right) = \dfrac{{11}}{{243}}$

Therefore, the probability that a candidate would get four or more correct answers just by guessing is $\dfrac{{11}}{{243}}$.

10. A Person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is $\dfrac{1}{{100}}$. What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice?

Ans: Assume that $X$ is a variable that represents the number of winning prizes in 50 lotteries. As the lotteries are repeatedly won, the trials are Bernoulli trials. 

It is known that 50 lotteries are bought by a person with a chance of winning the price is $\dfrac{1}{{100}}$.

It can be clearly said that the variable $X$ is found to have a binomial distribution with $n = 50$ and $p = \dfrac{1}{{100}}$.

Now,

As, $q = 1 - p$

$q = 1 - \dfrac{1}{{100}}$

$q = \dfrac{{99}}{{100}}$

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 50 for $n$, $\dfrac{1}{{100}}$ for $p$ and $\dfrac{{99}}{{100}}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^{50}}{C_x}{\left( {\dfrac{{99}}{{100}}} \right)^{50 - x}}{\left( {\dfrac{1}{{100}}} \right)^x}$

Now consider the conditions,

(a). The probability that he will  win a prize at least once.

$P\left( {{\text{Winning at least once}}} \right) = P\left( {X \geqslant 1} \right)$

$P\left( {X \geqslant 1} \right) = 1 - P\left( {X < 1} \right)$

$P\left( {X \geqslant 1} \right) = 1 - P\left( {X = 0} \right)$

$P\left( {X \geqslant 1} \right) = 1{ - ^{50}}{C_0}{\left( {\dfrac{{99}}{{100}}} \right)^{50}}$

On evaluating further,

$P\left( {X \geqslant 1} \right) = 1 - 1{\left( {\dfrac{{99}}{{100}}} \right)^{50}}$

$P\left( {X \geqslant 1} \right) = 1 - {\left( {\dfrac{{99}}{{100}}} \right)^{50}}$

Therefore, the probability that he will win a prize at least once is $1 - {\left( {\dfrac{{99}}{{100}}} \right)^{50}}$.

(b). The probability that he will win a prize exactly once.

$P\left( {{\text{Winning exactly once}}} \right) = P\left( {X = 1} \right)$

$P\left( {X = 1} \right){ = ^{50}}{C_1}{\left( {\dfrac{{99}}{{100}}} \right)^{49}}{\left( {\dfrac{1}{{100}}} \right)^1}$

On evaluating further,

$P\left( {X = 1} \right){ = ^{50}}{C_1}\left( {\dfrac{1}{{100}}} \right){\left( {\dfrac{{99}}{{100}}} \right)^{49}}$

It is known that $^50{C_1} = 50$,

$P\left( {X = 1} \right) = \left( {50} \right)\left( {\dfrac{1}{{100}}} \right){\left( {\dfrac{{99}}{{100}}} \right)^{49}}$

$P\left( {X = 1} \right) = \dfrac{1}{2}{\left( {\dfrac{{99}}{{100}}} \right)^{49}}$

Therefore, the probability that he will win a prize exactly once is $\dfrac{1}{2}{\left( {\dfrac{{99}}{{100}}} \right)^{49}}$.

(c). The probability that he will win a prize at least twice.

$P\left( {{\text{at least twice}}} \right) = P\left( {X \geqslant 2} \right)$

$P\left( {X \geqslant 2} \right) = 1 - P\left( {X < 2} \right)$

$P\left( {X \geqslant 2} \right) = 1 - P\left( {X \leqslant 1} \right)$

$P\left( {X \geqslant 2} \right) = 1 - \left[ {P\left( {X = 0} \right) + P\left( {X = 1} \right)} \right]$

$P\left( {X \geqslant 2} \right) = \left[ {1 - P\left( {X = 0} \right)} \right] - P\left( {X = 1} \right)$

Substituting the obtained values,

$P\left( {X \geqslant 2} \right) = 1 - {\left( {\dfrac{{99}}{{100}}} \right)^{50}} - \dfrac{1}{2}{\left( {\dfrac{{99}}{{100}}} \right)^{49}}$

On evaluating further,

$P\left( {X \geqslant 2} \right) = 1 - {\left( {\dfrac{{99}}{{100}}} \right)^{49}} \times \left[ {\dfrac{{99}}{{100}} + \dfrac{1}{2}} \right]$

$P\left( {X \geqslant 2} \right) = 1 - {\left( {\dfrac{{99}}{{100}}} \right)^{49}} \times \left[ {\dfrac{{149}}{{100}}} \right]$

$P\left( {X \geqslant 2} \right) = 1 - \left( {\dfrac{{149}}{{100}}} \right) \times {\left[ {\dfrac{{99}}{{100}}} \right]^{49}}$

Therefore, the probability that he will win a prize at least twice is  $1 - \left( {\dfrac{{149}}{{100}}} \right) \times {\left[ {\dfrac{{99}}{{100}}} \right]^{49}}$.

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

Ans: As the die is repeatedly tossed, these trials are Bernoulli trials. 

Consider the variable $X$ that represents the number of times of getting 5 in 7 throws of the die.

Probability of an experiment is supposed to be a ratio of number of favourable outcomes to the total number of outcomes. 

Thus, the probability of getting 5 in a single throw of die is $p = \dfrac{1}{6}$.

As, $q = 1 - p$

$q = 1 - \dfrac{1}{6}$

$q = \dfrac{5}{6}$

It is observed that $X$ is supposed to have a binomial distribution with $n = 7$ and $p = \dfrac{1}{6}$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 7 for $n$, $\dfrac{1}{6}$ for $p$ and $\dfrac{5}{6}$ for $q$ in the above formula.

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

$={ }^{7} C_{x}\left(\frac{5}{6}\right)^{7-x} \cdot\left(\frac{1}{6}\right)^{x}$

$\text { Probability of getting } 5 \text { exactly twice in 7 throws of a die }=\mathrm{P}(\mathrm{X}=2)$

$={ }^{7} C_{2}\left(\frac{5}{6}\right)^{5} \cdot\left(\frac{1}{6}\right)^{2}$

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

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

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

Ans: As the die is repeatedly tossed, these trials are Bernoulli trials. 

Consider the variable $X$ that represents the number of times of getting sixes in 6 throws of the die.

Probability of an experiment is supposed to be a ratio of number of favourable outcomes to the total number of outcomes. 

Thus, the probability of getting a six in a single throw of die is $p = \dfrac{1}{6}$.

As, $q = 1 - p$

$q = 1 - \dfrac{1}{6}$

$q = \dfrac{5}{6}$

It is observed that $X$ is supposed to have a binomial distribution with $n = 6$ and $p = \dfrac{1}{6}$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 6 for $n$, $\dfrac{1}{6}$ for $p$ and $\dfrac{5}{6}$ for $q$ in the above formula.

$P\left( {{\text{at most 2 sixes}}} \right) = P\left( {X \leqslant 2} \right)$

$P\left( {X \leqslant 2} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right)$

$P\left( {X \leqslant 2} \right){ = ^6}{C_0}{\left( {\dfrac{5}{6}} \right)^6}{ + ^6}{C_1}{\left( {\dfrac{5}{6}} \right)^5} \cdot \left( {\dfrac{1}{6}} \right){ + ^6}{C_2}{\left( {\dfrac{5}{6}} \right)^4} \cdot {\left( {\dfrac{1}{6}} \right)^2}$

$P\left( {X \leqslant 2} \right) = 1 \cdot {\left( {\dfrac{5}{6}} \right)^6} + 6{\left( {\dfrac{5}{6}} \right)^5} \cdot \left( {\dfrac{1}{6}} \right) + 15\left( {\dfrac{1}{{36}}} \right){\left( {\dfrac{5}{6}} \right)^4}$

$P\left( {X \leqslant 2} \right) = {\left( {\dfrac{5}{6}} \right)^4}\left[ {{{\left( {\dfrac{5}{6}} \right)}^2} + \left( {\dfrac{5}{6}} \right) + \left( {\dfrac{5}{{12}}} \right)} \right]$

On evaluating further,

$P\left( {X \leqslant 2} \right) = {\left( {\dfrac{5}{6}} \right)^4}\left[ {\dfrac{{25}}{{36}} + \dfrac{5}{6} + \dfrac{5}{{12}}} \right]$

$P\left( {X \leqslant 2} \right) = {\left( {\dfrac{5}{6}} \right)^4}\left[ {\dfrac{{25 + 30 + 15}}{{36}}} \right]$

$P\left( {X \leqslant 2} \right) = \left( {\dfrac{{70}}{{36}}} \right){\left( {\dfrac{5}{6}} \right)^4}$

$P\left( {X \leqslant 2} \right) = \left( {\dfrac{{35}}{{18}}} \right){\left( {\dfrac{5}{6}} \right)^4}$

Therefore, the probability of throwing at most 2 sixes in 6 throws of a single die is $\left( {\dfrac{{35}}{{18}}} \right){\left( {\dfrac{5}{6}} \right)^4}$.

13. It is known that $10\% $ of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?

Ans: The percentage of articles that are found to be defective is $10\% $. Repeated selections of these articles in a random sample space are Bernoulli trials.

Consider $X$ is a variable that denotes the number of defective articles out of a sample of 12 articles.

It is observed that $X$ is supposed to have a binomial distribution with $n = 12$ and $p = \dfrac{{10}}{{100}}$, that is $\dfrac{1}{{10}}$.

Now, $q = 1 - p$

$q = 1 - \dfrac{1}{{10}}$

$q = \dfrac{9}{{10}}$

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 12 for $n$, $\dfrac{1}{{10}}$ for $p$ and $\dfrac{9}{{10}}$ for $q$ in the above formula.

$P\left( {X = x} \right){ = ^{12}}{C_x}{\left( {\dfrac{9}{{10}}} \right)^{12 - x}} \cdot {\left( {\dfrac{1}{{10}}} \right)^x}$

Consider, 

$P\left( {{\text{Selecting 9 defective articles}}} \right) = P\left( {X = 9} \right)$

$P\left( {X = 9} \right){ = ^{12}}{C_9}{\left( {\dfrac{9}{{10}}} \right)^{12 - 9}}{\left( {\dfrac{1}{{10}}} \right)^9}$

It is known that, $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

$P\left( {X = 9} \right) = \left( {\dfrac{{12!}}{{3!9!}}} \right){\left( {\dfrac{9}{{10}}} \right)^{12 - 9}}{\left( {\dfrac{1}{{10}}} \right)^9}$

$P\left( {X = 9} \right) = \left( {220} \right){\left( {\dfrac{9}{{10}}} \right)^3}{\left( {\dfrac{1}{{10}}} \right)^9}$

$P\left( {X = 9} \right) = \dfrac{{22 \times {9^3}}}{{{{10}^{11}}}}$

Therefore, the probability that in a random sample of 12 such articles, 9 are defective is $\dfrac{{22 \times {9^3}}}{{{{10}^{11}}}}$.

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

1. ${10^{ - 1}}$

2. ${\left( {\dfrac{1}{2}} \right)^5}$

3. \[{\left( {\dfrac{9}{{10}}} \right)^5}\]

4. $\dfrac{9}{{10}}$

Ans: It is given that 10 bulbs out of 100 bulbs in a box are defective. As the selection of the bulbs is repeated, these are Bernoulli trials.

Consider $X$ is a variable that denotes the number of defective bulbs out of a sample of 5 bulbs.

${\text{Probablity of getting a defective bulb, }}p = \dfrac{{10}}{{100}}$

$p = \dfrac{1}{{10}}$

Now, it is known that $q = 1 - p$.

$q = 1 - \dfrac{1}{{10}}$

$q = \dfrac{9}{{10}}$

Thus, it is observed that $X$ is supposed to have a binomial distribution with $n = 5$ and $p = \dfrac{1}{{10}}$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 5 for $n$, $\dfrac{1}{{10}}$ for $p$ and $\dfrac{9}{{10}}$ for $q$ in the above formula.

$P\left( {{\text{none of the bulb is defective}}} \right) = P\left( {X = 0} \right)$

$P\left( {X = 0} \right){ = ^5}{C_0}{\left( {\dfrac{9}{{10}}} \right)^{5 - 0}}{\left( {\dfrac{1}{{10}}} \right)^0}$

$P\left( {X = 0} \right) = \left( 1 \right){\left( {\dfrac{9}{{10}}} \right)^5}$

$P\left( {X = 0} \right) = {\left( {\dfrac{9}{{10}}} \right)^5}$

Therefore, the correct option is found to be (C), ${\left( {\dfrac{9}{{10}}} \right)^5}$.

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

1. $^5{C_4}{\left( {\dfrac{4}{5}} \right)^4}\dfrac{1}{5}$

2. ${\left( {\dfrac{4}{5}} \right)^4}\dfrac{1}{5}$

3. $^5{C_1}\dfrac{1}{5}{\left( {\dfrac{4}{5}} \right)^4}$

4. None of these

Ans: It is given that the probability that the student is not swimmer is $\dfrac{1}{5}$. As the selection of the students who are swimmers is repeated, these are Bernoulli trials.

Consider $X$ to be a variable that denotes the number of students out of 5 students who are swimmers.

The probability that the student is not a swimmer is $\dfrac{1}{5}$, that is, $q = \dfrac{1}{5}$.

It is known that, $p = 1 - q$.

Thus,

$p = 1 - \dfrac{1}{5}$

$p = \dfrac{4}{5}$

Thus, it is observed that $X$ is supposed to have a binomial distribution with $n = 5$ and $p = \dfrac{4}{5}$.

Now, the probability is calculated with the help of the formula, 

$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$

Substitute 5 for $n$, $\dfrac{4}{5}$ for $p$ and $\dfrac{1}{5}$ for $q$ in the above formula.

$P\left( {{\text{four students are swimmers}}} \right) = P\left( {X = 4} \right)$

$P\left( {X = 4} \right){ = ^5}{C_4}{\left( {\dfrac{1}{5}} \right)^1}{\left( {\dfrac{4}{5}} \right)^4}$

Therefore, the correct option is found to be (A), $P\left( {X = 4} \right){ = ^5}{C_4}{\left( {\dfrac{4}{5}} \right)^4}\left( {\dfrac{1}{5}} \right)$.

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.5

Opting for the NCERT solutions for Ex 13.5 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.5 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 13 Exercise 13.5 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 13 Exercise 13.5, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 13 Exercise 13.5 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.