# NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.5

## Chapter 13 Ex.13.5 Question 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?

### Solution

Let X denote the number of success of getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is, $$p = \frac{3}{6} = \frac{1}{2}$$

$$\therefore \;\;q = 1 - p = \frac{1}{2}$$

X has a binominal distribution.

Thus, $${\text{P}}\left( {{\text{X}} = {\text{x}}} \right) = {}^n{{\text{C}}_x}{q^{n - x}}{p^x},{\text{where n}} = {\text{ 0, 1, 2 }}....n$$

\begin{align} &= {}^6{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^{6 - x}} \times {\left( {\frac{1}{2}} \right)^x}\\ &= {}^6{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^6}\end{align}

(i) $${\text{P}}\left( {{\text{5 success}}} \right) = {\text{P}}\left( {{\text{X}} = 5} \right)$$

\begin{align} &= {}^6{{\text{C}}_5}{\left( {\frac{1}{2}} \right)^6}\\ &= 6 \times \frac{1}{{64}} = \frac{3}{{32}}\end{align}

(ii) $${\text{P}}\left( {{\text{at least 5 success}}} \right) = {\text{P}}\left( {{\text{X}} \ge 5} \right)$$

\begin{align} &= {\text{P}}\left( {{\text{X}} = 5} \right) + {\text{P}}\left( {{\text{X}} = 6} \right)\\ &= {}^6{{\text{C}}_5}{\left( {\frac{1}{2}} \right)^6} + {}^6{{\text{C}}_6}{\left( {\frac{1}{2}} \right)^6}\\ &= 6 \times \frac{1}{{64}} + 1 \times \frac{1}{{64}} = \frac{7}{{64}}\end{align}

(iii) $${\text{P}}\left( {{\text{at most 5 success}}} \right) = {\text{P}}\left( {{\text{X}} \le 5} \right)$$

\begin{align} &= 1 - {\text{P}}\left( {{\text{X}} > 5} \right)\\&{\text{ = }}1 - {\text{P}}\left( {{\text{X}} = 6} \right)\\& = 1 - {}^6{{\text{C}}_6}{\left( {\frac{1}{2}} \right)^6}\\ &= 1 - \frac{1}{{64}} = \frac{{63}}{{64}}\end{align}

## Chapter 13 Ex.13.5 Question 2

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

### Solution

Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is

\begin{align}p &= \frac{6}{{36}} = \frac{1}{6}\\&\therefore \;\;q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}\end{align}

Clearly, X has the binomial distribution with 

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x},{\text{where }}x{\text{ = 0, 1, 2, 3 }}....n\\& = {}^4{{\text{C}}_x}{\left( {\frac{5}{6}} \right)^{4 - x}} \times {\left( {\frac{1}{6}} \right)^x}\\ &= {}^4{{\text{C}}_x} \times \frac{{{5^{4 - x}}}}{{{6^4}}}\\&\therefore \;\;{\text{P}}\left( {{\text{2 successes}}} \right) = {\text{P}}\left( {{\text{X}} = 2} \right)\\ &= {}^4{{\text{C}}_2} \times \frac{{{5^{4 - 2}}}}{{{6^4}}}\\& = 6 \times \frac{{25}}{{1296}} = \frac{{25}}{{216}}\end{align}

## Chapter 13 Ex.13.5 Question 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?

### Solution

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

\begin{align} &\Rightarrow \; p = \frac{5}{{100}} = \frac{1}{{20}}\\&\therefore \;\;q = 1 - \frac{1}{{20}} = \frac{{19}}{{20}}\end{align}

X has a binomial distribution with$$n = 10,\;p = \frac{1}{{20}}\;{\text{and}}\;q = \frac{{19}}{{20}}$$

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x},{\text{where }}x{\text{ = 0, 1, 2 }}....n\\& = {}^{10}{{\text{C}}_x}{\left( {\frac{{19}}{{20}}} \right)^{10 - x}} \times {\left( {\frac{1}{{20}}} \right)^x}\\\\{\text{P}}\left( {{\text{not more than 1 defective item}}} \right) &= {\text{P}}\left( {{\text{X}} \le 1} \right)\\[5pt] &= {\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)\\&= {}^{10}{{\text{C}}_0}{\left( {\frac{{19}}{{20}}} \right)^{10}} \times {\left( {\frac{1}{{20}}} \right)^0} + {}^{10}{{\text{C}}_1}{\left( {\frac{{19}}{{20}}} \right)^9} \times {\left( {\frac{1}{{20}}} \right)^1}\\ &= {\left( {\frac{{19}}{{20}}} \right)^{10}} + 10 \times {\left( {\frac{{19}}{{20}}} \right)^9} \times \left( {\frac{1}{{20}}} \right)\\ &= {\left( {\frac{{19}}{{20}}} \right)^9}\left( {\frac{{19}}{{20}} + \frac{{10}}{{20}}} \right)\\ &= \left( {\frac{{29}}{{20}}} \right) \times {\left( {\frac{{19}}{{20}}} \right)^9}\end{align}

## Chapter 13 Ex.13.5 Question 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?

### Solution

Let X represent the number of spade cards among the five cards drawn.

In a well shuffled deck of 52 cards, there are 13 spades cards.

\begin{align}p& = \frac{{13}}{{52}} = \frac{1}{4}\\\therefore \;\;q &= 1 - \frac{1}{4} = \frac{3}{4}\end{align}

X has a binomial distribution with $$n = 5,\;p = \frac{1}{4}\;{\text{and}}\;q = \frac{3}{4}$$

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x},{\text{where }}x{\text{ = 0, 1, 2 }}....n\\ &= {}^5{{\text{C}}_x}{\left( {\frac{3}{4}} \right)^{5 - x}} \times {\left( {\frac{1}{4}} \right)^x}\end{align}

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

\begin{align} &= {}^5{{\text{C}}_5}{\left( {\frac{3}{4}} \right)^0} \times {\left( {\frac{1}{4}} \right)^5}\\ &= 1 \times \frac{1}{{1024}} = \frac{1}{{1024}}\end{align}

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

\begin{align} &= {}^5{{\text{C}}_3}{\left( {\frac{3}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^3}\\ &= 10 \times \frac{9}{{16}} \times \frac{1}{{64}}\\ &= \frac{{45}}{{512}}\end{align}

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

\begin{align} &= {}^5{{\text{C}}_0}{\left( {\frac{3}{4}} \right)^5} \times {\left( {\frac{1}{4}} \right)^0}\\ &= 1 \times \frac{{243}}{{1024}} = \frac{{243}}{{1024}}\end{align}

## Chapter 13 Ex.13.5 Question 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

(i) none

(ii) not more than one

(iii) more than one

(iv) at least one

Will fuse after 150 days of use.

### Solution

Let X represents the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials.

Given, $$p = 0.05$$

$$\therefore \;\;q = 1 - p = 1 - 0.05 = 0.95$$

X has a binomial distribution with $$n = 5,\;p = 0.05\;{\text{and}}\;q = 0.95$$

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right)&= {}^n{{\text{C}}_x}{q^{n - x}}{p^x},{\text{where }}x{\text{ = 0, 1, 2 }}....n\\ &= {}^5{{\text{C}}_x} \times {\left( {0.95} \right)^{5 - x}} \times {\left( {0.05} \right)^x}\end{align}

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

\begin{align} &= {}^5{{\text{C}}_0} \times {\left( {0.95} \right)^5} \times {\left( {0.05} \right)^0}\\ &= 1 \times {\left( {0.95} \right)^5}\\ &= {\left( {0.95} \right)^5}\\ &= 0.7737\end{align}

(ii)$${\text{P}}\left( {{\text{not more than one}}} \right) = {\text{P}}\left( {{\text{X}} \le 1} \right)$$

\begin{align} &= {\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)\\ &= {}^5{{\text{C}}_0} \times {\left( {0.95} \right)^5} \times {\left( {0.05} \right)^0} + {}^5{{\text{C}}_1} \times {\left( {0.95} \right)^4} \times {\left( {0.05} \right)^1}\\ &= 1 \times {\left( {0.95} \right)^5} + 5 \times {\left( {0.95} \right)^4} \times \left( {0.05} \right)\\& = {\left( {0.95} \right)^5} + \left( {0.25} \right) \times {\left( {0.95} \right)^4}\\& = {\left( {0.95} \right)^4} \times \left[ {0.95 + 0.25} \right]\\& = {\left( {0.95} \right)^4} \times 1.2\\ &= 0.977\end{align}

(iii)$${\text{P}}\left( {{\text{more than 1}}} \right) = {\text{P}}\left( {{\text{X}} > 1} \right)$$

\begin{align} = 1 - {\text{P}}\left( {{\text{X}} \le 1} \right)\\ = 1 - {\text{P}}\left( {{\text{not more than 1}}} \right)\\ = 1 - {\left( {0.95} \right)^4} \times 1.2\\ = 0.02\end{align}

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

\begin{align} &= 1 - {\text{P}}\left( {{\text{X}} < 1} \right)\\ &= 1 - {\text{P}}\left( {{\text{X}} = 0} \right)\\ &= 1 - {}^5{{\text{C}}_0} \times {\left( {0.95} \right)^5} \times {\left( {0.05} \right)^0}\\ &= 1 - 1 \times {\left( {0.95} \right)^5}\\& = 1 - {\left( {0.95} \right)^5}\\ &= 0.2263\end{align}

## Chapter 13 Ex.13.5 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?

### Solution

Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.

X has a binominal distribution with $$n = 4\;{\text{and}}\;p = \frac{1}{{10}}\;$$

$$\therefore \;\;q = 1 - p = 1 - \frac{1}{{10}} = \frac{9}{{10}}$$

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x},{\text{where }}x = {\text{1, 2 }}....n\\ &= {}^4{{\text{C}}_x}{\left( {\frac{9}{{10}}} \right)^{4 - x}} \times {\left( {\frac{1}{{10}}} \right)^x}\\\\{\text{P}}\left( {{\text{none marked with 0}}} \right)& = {\text{P}}\left( {{\text{X}} = 0} \right)\\& = {}^4{{\text{C}}_0}{\left( {\frac{9}{{10}}} \right)^4} \times {\left( {\frac{1}{4}} \right)^0}\\ &= 1 \times {\left( {\frac{9}{{10}}} \right)^4} = {\left( {\frac{9}{{10}}} \right)^4}\end{align}

## Chapter 13 Ex.13.5 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 tail, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

### Solution

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

\begin{align}\therefore \;p &= \frac{1}{2}\;\\\therefore \;\;q &= 1 - p = 1 - \frac{1}{2} = \frac{1}{2}\end{align}

X has a binominal distribution with $$n = 20\;{\text{and}}\;p = \frac{1}{2}\;$$

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x},{\text{where }}x = 0,{\text{1, 2 }}....n\\ &= {}^{20}{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^{20 - x}} \times {\left( {\frac{1}{2}} \right)^x}\\& = {}^{20}{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^{20}}\end{align}

\begin{align}{\text{P}}\left( {{\text{at least 12 questions answered correctly}}} \right) &= {\text{P}}\left( {{\text{X}} \ge 12} \right)\\ &= {\text{P}}\left( {{\text{X}} = 12} \right) + {\text{P}}\left( {{\text{X}} = 13} \right) + ... + {\text{P}}\left( {{\text{X}} = 20} \right)\\ &= {}^{20}{{\text{C}}_{12}}{\left( {\frac{1}{2}} \right)^{20}} + {}^{20}{{\text{C}}_{13}}{\left( {\frac{1}{2}} \right)^{20}} + ... + {}^{20}{{\text{C}}_{20}}{\left( {\frac{1}{2}} \right)^{20}}\\ &= {\left( {\frac{1}{2}} \right)^{20}} \times \left[ {{}^{20}{{\text{C}}_{12}} + {}^{20}{{\text{C}}_{13}} + ... + {}^{20}{{\text{C}}_{20}}} \right]\end{align}

## Chapter 13 Ex.13.5 Question 8

Suppose X has a binominal distribution $${\text{B}}\left( {6 \times \frac{1}{2}} \right)$$ . Show that $${\text{X}} = 3$$ is the most likely outcome.

(Hint: $${\text{P}}\left( {\text{X}} \right) = 3$$ is the maximum among all $${\text{P}}\left( {{{\text{x}}_{\text{i}}}} \right),\;{{\text{x}}_{\text{i}}} = 0,1,2,3,4,5,6$$)

### Solution

X is the random variable whose binomial distribution is $${\text{B}}\left( {6,\frac{1}{2}} \right)$$

Thus, $$n = 6\;{\text{and}}\;p = \frac{1}{2}\;$$

$$\therefore \;\;q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$$

Then,

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x}\\ &= {}^6{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^{6 - x}} \times {\left( {\frac{1}{2}} \right)^x}\\ &= {}^6{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^6}\end{align}

It can be seen that $${\text{P}}\left( {{\text{X}} = x} \right)$$ will be maximum, if $${}^6{{\text{C}}_x}$$ will be maximum. $${\text{Then, }}{}^6{{\text{C}}_0} = {}^6{{\text{C}}_6} = \frac{{6!}}{{0!\; \times 6!}} = 1$$

\begin{align}{}^6{{\text{C}}_1} = {}^6{{\text{C}}_5} &= \frac{{6!}}{{1!\; \times 5!}} = 6\\{}^6{{\text{C}}_2} = {}^6{{\text{C}}_4} &= \frac{{6!}}{{2!\; \times 4!}} = 15\\{}^6{{\text{C}}_3}& = \frac{{6!}}{{3!\; \times 3!}} = 20\end{align}

The value of $${}^6{{\text{C}}_3}$$ is maximum. Therefore, for $$x~=\text{ 3},\text{ P}\left( \text{X=x} \right)$$ is maximum.

Thus, $${\text{X}} = {\text{3}}$$ is the most likely outcome.

## Chapter 13 Ex.13.5 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?

### Solution

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

Probability of getting a correct answer is, $$p = \frac{1}{3}\;$$

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

Clearly, X has a binomial distribution with $$n = 5\;{\text{and}}\;p = \frac{1}{3}\;$$

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x}\\ &= {}^5{{\text{C}}_x}{\left( {\frac{2}{3}} \right)^{5 - x}} \times {\left( {\frac{1}{3}} \right)^x}\end{align}

\begin{align}{\text{P}}\left( {{\text{guessing more than 4 correct answers}}} \right) &= {\text{P}}\left( {{\text{X}} \ge 4} \right)\\[5pt] &= {\text{P}}\left( {{\text{X}} = 4} \right) + {\text{P}}\left( {{\text{X}} = 5} \right)\\ &= {}^5{{\text{C}}_4}\left( {\frac{2}{3}} \right) \times {\left( {\frac{1}{3}} \right)^4} + {}^5{{\text{C}}_5} \times {\left( {\frac{1}{3}} \right)^5}\\& = 5 \times \frac{2}{3} \times \frac{1}{{81}} + 1 \times \frac{1}{{243}}\\ &= \frac{{10}}{{243}} + \frac{1}{{243}} = \frac{{11}}{{243}}\end{align}

## Chapter 13 Ex.13.5 Question 10

A person buys a lottery ticket in $$50$$ lotteries, in each of which his change of winning a prize is $$\frac{1}{{100}}.$$ What is the probability that he will in a prize

(A) at least once

(B) exactly once

(C) at least twice?

### Solution

Let $$X$$ represents the number of winning prizes in $$50$$ lotteries.

Clearly, $$X$$ has a binominal distribution with $$n = 50\;{\text{and}}\;p = \frac{1}{{100}}\;$$

$$\therefore \;\;q = 1 - p = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}$$

\begin{align}{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{q^{n - x}}{p^x}\\ &= {}^{50}{{\text{C}}_x}{\left( {\frac{{99}}{{100}}} \right)^{50 - x}} \times {\left( {\frac{1}{{100}}} \right)^x}\end{align}

(a) $${\text{P}}\left( {{\text{winning at least once}}} \right) = {\text{P}}\left( {{\text{X}} \ge 1} \right)$$

\begin{align} &= 1 - {\text{P}}\left( {{\text{X}} < 1} \right) \\[5pt]&= 1 - {\text{P}}\left( {{\text{X}} = 0} \right)\\ &= 1 - {}^{50}{{\text{C}}_0} \times {\left( {\frac{{99}}{{100}}} \right)^{50}}\end{align}

\begin{align} &= 1 - 1 \times {\left( {\frac{{99}}{{100}}} \right)^{50}}\\ &= 1 - {\left( {\frac{{99}}{{100}}} \right)^{50}}\end{align}

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

\begin{align} &= {}^{50}{{\text{C}}_1} \times {\left( {\frac{{99}}{{100}}} \right)^{49}} \times {\left( {\frac{1}{{100}}} \right)^1}\\& = 50 \times \left( {\frac{1}{{100}}} \right) \times {\left( {\frac{{99}}{{100}}} \right)^{49}}\\ &= \frac{1}{2} \times {\left( {\frac{{99}}{{100}}} \right)^{49}}\end{align}

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

\begin{align} &= 1 - {\text{P}}\left( {{\text{X}} < 2} \right)\\[5pt]&= 1 - {\text{P}}\left( {{\text{X}} \le 1} \right)\\[5pt]&= 1 - \left[ {{\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)} \right]\\[5pt]&= \left[ {1 - {\text{P}}\left( {{\text{X}} = 0} \right)} \right] - {\text{P}}\left( {{\text{X}} = 1} \right)\\ &= 1 - {\left( {\frac{{99}}{{100}}} \right)^{50}} - \frac{1}{2} \times {\left( {\frac{{99}}{{100}}} \right)^{49}}\\ &= 1 - {\left( {\frac{{99}}{{100}}} \right)^{49}} \times \left[ {\frac{{99}}{{100}} + \frac{1}{2}} \right]\\ &= 1 - {\left( {\frac{{99}}{{100}}} \right)^{49}} \times \left( {\frac{{149}}{{100}}} \right)\\ &= 1 - \left( {\frac{{149}}{{100}}} \right) \times {\left( {\frac{{99}}{{100}}} \right)^{49}}\end{align}

## Chapter 13 Ex.13.5 Question 11

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

### Solution

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

Probability of getting 5 in a single throw of the die, $$p = \frac{1}{6}\;$$

$$\therefore \;\;q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$$

Clearly, X has the probability distribution with $$n = 7\;{\text{and}}\;p = \frac{1}{6}\;$$

$$\therefore {\text{ P}}\left( {{\text{X}} = x} \right) = {}^n{{\text{C}}_x}{q^{n - x}}{p^x} = {}^7{{\text{C}}_x}{\left( {\frac{5}{6}} \right)^{7 - x}} \times {\left( {\frac{1}{6}} \right)^x}$$

\begin{align}{\text{P}}\left( {{\text{getting 5 exactly twice}}} \right) &= {\text{P}}\left( {{\text{X}} = 2} \right)\\ &= {}^7{{\text{C}}_2} \times {\left( {\frac{5}{6}} \right)^5} \times {\left( {\frac{1}{6}} \right)^2}\\ &= 21 \times {\left( {\frac{5}{6}} \right)^5} \times \frac{1}{{36}}\\& = \frac{7}{{12}} \times {\left( {\frac{5}{6}} \right)^5}\end{align}

## Chapter 13 Ex.13.5 Question 12

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

### Solution

Let X represent the number of times of getting sixes in 6 throws of the die.

Probability of getting six in a single throw of die, $$p = \frac{1}{6}\;$$

$$\therefore \;\;q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$$

Clearly, X has a binomial distribution with $$n = 6$$

$$\therefore {\text{ P}}\left( {{\text{X}} = x} \right) = {}^n{{\text{C}}_x}{q^{n - x}}{p^x} = {}^6{{\text{C}}_x}{\left( {\frac{5}{6}} \right)^{6 - x}} \times {\left( {\frac{1}{6}} \right)^x}$$

\begin{align}{\text{P}}\left( {{\text{at most 2 sixes}}} \right)& = {\text{P}}\left( {{\text{X}} \le 2} \right)\\[5pt]&= {\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right) + {\text{P}}\left( {{\text{X}} = 2} \right)\\ &= {}^6{{\text{C}}_0}{\left( {\frac{5}{6}} \right)^6} + {}^6{{\text{C}}_1}{\left( {\frac{5}{6}} \right)^5} \times \left( {\frac{1}{6}} \right) + {}^6{{\text{C}}_2}{\left( {\frac{5}{6}} \right)^4} \times {\left( {\frac{1}{6}} \right)^2}\\ &= 1 \times {\left( {\frac{5}{6}} \right)^6} + 6 \times \frac{1}{6} \times {\left( {\frac{5}{6}} \right)^5} + 15 \times \frac{1}{{36}} \times {\left( {\frac{5}{6}} \right)^4}\\&= {\left( {\frac{5}{6}} \right)^6} + {\left( {\frac{5}{6}} \right)^5} + \frac{5}{{12}} \times {\left( {\frac{5}{6}} \right)^4}\\ &= {\left( {\frac{5}{6}} \right)^4}\left[ {{{\left( {\frac{5}{6}} \right)}^2} + \left( {\frac{5}{6}} \right) + \left( {\frac{5}{{12}}} \right)} \right]\\ &= {\left( {\frac{5}{6}} \right)^4}\left[ {\frac{{25}}{{36}} + \frac{5}{6} + \frac{5}{{12}}} \right]\\& = {\left( {\frac{5}{6}} \right)^4}\left[ {\frac{{25 + 30 + 15}}{{36}}} \right] \\&= \frac{{70}}{{36}} \times {\left( {\frac{5}{6}} \right)^4} = \frac{{35}}{{18}} \times {\left( {\frac{5}{6}} \right)^4}\\&= 0.9377\end{align}

## Chapter 13 Ex.13.5 Question 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?

### Solution

Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.

Clearly, X has a binominal distribution with $$n = 12{\text{ and }}p = 10\% = \frac{{10}}{{100}} = \;\frac{1}{{10}}$$

$$\therefore \;\;q = 1 - p = 1 - \frac{1}{{10}} = \frac{9}{{10}}$$

$$\therefore {\text{ P}}\left( {{\text{X}} = x} \right) = {}^n{{\text{C}}_x}{q^{n - x}}{p^x} = {}^{12}{{\text{C}}_x}{\left( {\frac{9}{{10}}} \right)^{12 - x}} \times {\left( {\frac{1}{{10}}} \right)^x}$$

\begin{align}{\text{P}}\left( {{\text{Selecting 9 defective articles}}} \right) &= {}^{12}{{\text{C}}_9} \times {\left( {\frac{9}{{10}}} \right)^3} \times {\left( {\frac{1}{{10}}} \right)^9}\\&= 220 \times \frac{{{9^3}}}{{{{10}^3}}} \times \frac{1}{{{{10}^9}}}\\ &= \frac{{22 \times {9^3}}}{{{{10}^{11}}}}\end{align}

## Chapter 13 Ex.13.5 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

\begin{align}&\left( {\text{A}} \right)\;{10^{ - 1}}\\&\left( {\text{B}} \right)\;{\left( {\frac{1}{2}} \right)^5}\\&\left( {\text{C}} \right)\;{\left( {\frac{9}{{10}}} \right)^5}\\&\left( {\text{D}} \right)\;\frac{9}{{10}}\end{align}

### Solution

Let X denote the number of defective bulbs out of a sample of 5 bulbs.

Probability of getting a defective bulb $$p = \frac{{10}}{{100}} = \;\frac{1}{{10}}$$

$$\therefore \;\;q = 1 - p = 1 - \frac{1}{{10}} = \frac{9}{{10}}$$

Clearly, X has binominal distribution with $$n = 5{\text{ and }}p = \;\frac{1}{{10}}$$

$$\therefore {\text{ P}}\left( {{\text{X}} = x} \right) = {}^n{{\text{C}}_x}{q^{n - x}}{p^x} = {}^5{{\text{C}}_x}{\left( {\frac{9}{{10}}} \right)^{5 - x}} \times {\left( {\frac{1}{{10}}} \right)^x}$$

\begin{align}{\text{P}}\left( {{\text{none of the bulbs is defective}}} \right)& = {\text{P}}\left( {{\text{X}} = 0} \right)\\ &= {}^5{{\text{C}}_0} \times {\left( {\frac{9}{{10}}} \right)^5}\\ &= 1 \times {\left( {\frac{9}{{10}}} \right)^5}\\& = {\left( {\frac{9}{{10}}} \right)^5}\end{align}

The probability that a student is not a swimming is $$\frac{1}{5}$$. The probability that out of five students, four are swimmers is