# NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.4

## Chapter 13 Ex.13.4 Question 1

State which of the following are not the probability distribution of a random variable. Give reasons for your answer.

(i)

 X $$0$$ $$1$$ $$2$$ P (X) $$0.4$$ $$0.4$$ $$0.2$$

(ii)

 X $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ P (X) $$0.1$$ $$0.5$$ $$0.2$$ $$-0.1$$ $$0.3$$

(iii)

 Y $$-1$$ $$0$$ $$1$$ P (Y) $$0.6$$ $$0.1$$ $$0.2$$

(iv)

 Z $$3$$ $$2$$ $$1$$ $$0$$ $$-1$$ P (Z) $$0.3$$ $$0.2$$ $$0.4$$ $$0.1$$ $$0.05$$

### Solution

Since the sum of all the probabilities in a probability distribution is one.

(i) Sum of the probabilities$$= 0.4 + 0.4 + 0.2 = 1{\text{ }}$$

Thus, the given table is a probability distribution of random variable.

(ii) For $${\text{X}} = {\text{3}},{\text{ P}}\left( {\text{X}} \right) = - 0.1{\text{ }}$$

Since probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.

(iii) Sum of the probabilities$$= 0.6 + 0.1 + 0.2 = 0.9 \ne 1{\text{ }}$$

Thus, the given table is not a probability distribution of random variables.

(iv) Sum of the probabilities$$= 0.{\text{3}} + 0.{\text{2}} + 0.{\text{4}} + 0.{\text{1}} + 0.0{\text{5}} = {\text{1}}.0{\text{5}} \ne {\text{1}}$$

Thus, the given table is not a probability distribution of random variable.

## Chapter 13 Ex.13.4 Question 2

An urn contains $$5$$ red and $$2$$ black balls. Two balls are randomly drawn. Let X$$X$$ represents the number of black balls. What are the possible values of $$X$$? Is $$X$$ a random variable?

### Solution

Let B represents a black ball and R represents a red ball.

The two balls selected can be represented as $$BB, BR, RB, RR$$.

\begin{align}{\text{X}}\left( {{\text{BB}}} \right)&{\text{ = 2}}\\{\text{X}}\left( {{\text{BR}}} \right)&{\text{ = 1}}\\{\text{X}}\left( {{\text{RB}}} \right)&{\text{ = 1}}\\{\text{X}}\left( {{\text{RR}}} \right)&{\text{ = 0}}\end{align}

Thus, the possible values of X are $$0, 1,$$ and $$2$$.

Yes, $$X$$ is a random variable.

## Chapter 13 Ex.13.4 Question 3

Let $$X$$ represents the difference between the number of heads and the number of tails obtained when a coin is tossed $$6$$ times. What are possible values of $$X$$?

### Solution

A coin is tossed six times and $$X$$ represents the difference between the number of heads and the number of tails.

\begin{align}{\text{X}}\left( {{\text{6H, 0T}}} \right) = \left| {6 - 0} \right| = 6\\{\text{X}}\left( {{\text{5H, 1T}}} \right) = \left| {5 - 1} \right| = 4\\{\text{X}}\left( {{\text{4H, 2T}}} \right) = \left| {4 - 2} \right| = 2\\{\text{X}}\left( {{\text{3H, 3T}}} \right) = \left| {3 - 3} \right| = 0\\{\text{X}}\left( {{\text{2H, 4T}}} \right) = \left| {2 - 4} \right| = 2\\{\text{X}}\left( {{\text{1H, 5T}}} \right) = \left| {1 - 5} \right| = 4\\{\text{X}}\left( {{\text{0H, 6T}}} \right) = \left| {0 - 6} \right| = 6\end{align}

Therefore, the possible values of $$X$$ are $$6, 4, 2$$, and $$0$$.

## Chapter 13 Ex.13.4 Question 4

Find the probability distribution of

(i) number of heads in two tosses of a coin

(ii) number of tails in the simultaneous tosses of three coins

(iii) number of heads in four tosses of a coin

### Solution

(i) When one coin is tossed twice. the sample space is $$\left\{ {{\text{HH, HT, TH, TT}}} \right\}$$

Let X represent the number of heads.

$$\therefore {\text{X}}\left( {{\text{HH}}} \right) = {\text{2,}}\;{\text{X}}\left( {{\text{HT}}} \right) = {\text{1,}}\;{\text{X}}\left( {{\text{TH}}} \right) = {\text{1,}}\;{\text{X}}\left( {{\text{TT}}} \right) = 0$$

Thus, X can take the value of $$0, 1,$$ or $$2$$.

Since,

$${\text{P}}\left( {{\text{HH}}} \right) = {\text{P}}\left( {{\text{HT}}} \right) = {\text{P}}\left( {{\text{TH}}} \right) = {\text{P}}\left( {{\text{TT}}} \right) = \frac{1}{4}$$

\begin{align}{\text{P}}\left( {{\text{X}} = 0} \right)& = {\text{P}}\left( {{\text{TT}}} \right) = \frac{1}{4}\\{\text{P}}\left( {{\text{X}} = 1} \right) & {\text{P}}\left( {{\text{HT}}} \right) + {\text{P}}\left( {{\text{TH}}} \right) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\\{\text{P}}\left( {{\text{X}} = 2} \right) &= {\text{P}}\left( {{\text{HH}}} \right) = \frac{1}{4}\end{align}

Therefore, the required probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ P (X) $$\frac{1}{4}$$ $$\frac{1}{2}$$ $$\frac{1}{4}$$

(ii) The sample space is $$\left\{ {{\text{HHH,}}\;{\text{HHT,}}\;{\text{HTH,}}\;{\text{HTT,}}\;{\text{THH,}}\;{\text{THT,}}\;{\text{TTH,}}\;{\text{TTT}}} \right\}$$

Let X represents the number of tails.

Since, X can take the value of $$0, 1, 2,$$ or $$3$$.

\begin{align}{\text{P}}\left( {{\text{X}} = 0} \right) &= {\text{P}}\left( {{\text{HHH}}} \right) = \frac{1}{8}\\{\text{P}}\left( {{\text{X}} = 1} \right) &= {\text{P}}\left( {{\text{HHT}}} \right) + {\text{P}}\left( {{\text{HTH}}} \right) + {\text{P}}\left( {{\text{THH}}} \right) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}\\{\text{P}}\left( {{\text{X}} = 2} \right) &= {\text{P}}\left( {{\text{HTT}}} \right) + {\text{P}}\left( {{\text{THT}}} \right) + {\text{P}}\left( {{\text{TTH}}} \right) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}\\{\text{P}}\left( {{\text{X}} = 3} \right) &= {\text{P}}\left( {{\text{TTT}}} \right) = \frac{1}{8}\end{align}

Therefore, the probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ $$3$$ P (X) $$\frac{1}{8}$$ $$\frac{3}{8}$$ $$\frac{3}{8}$$ $$\frac{1}{8}$$

(iii) The sample space is {\text{S}} = \left\{ \begin{align}{\text{HHHH,}}\;{\text{HHHT,}}\;{\text{HHTH,}}\;{\text{HHTT,}}\;{\text{HTHT,}}\;{\text{HTHH,}}\;{\text{HTTH,}}\;{\text{HTTT}}\\{\text{THHH,}}\;{\text{THHT,}}\;{\text{THTH,}}\;{\text{THTT,}}\;{\text{TTHH,}}\;{\text{TTHT,}}\;{\text{TTTH,}}\;{\text{TTTT}}\end{align} \right\}

Let X be the random variable, which represents the number of heads.

Since, X can take the value of $$0, 1, 2, 3,$$or $$4$$.

\begin{align}{\text{P}}\left( {{\text{X}} = 0} \right)& = {\text{P}}\left( {{\text{TTTT}}} \right) = \frac{1}{{16}}\\\\{\text{P}}\left( {{\text{X}} = 1} \right)& = {\text{P}}\left( {{\text{TTTH}}} \right) + {\text{P}}\left( {{\text{TTHT}}} \right) + {\text{P}}\left( {{\text{THTT}}} \right) + {\text{P}}\left( {{\text{HTTT}}} \right)\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} = \frac{4}{{16}} = \frac{1}{4}\\\\{\text{P}}\left( {{\text{X}} = 2} \right)& = {\text{P}}\left( {{\text{HHTT}}} \right) + {\text{P}}\left( {{\text{THHT}}} \right) + {\text{P}}\left( {{\text{TTHH}}} \right){\text{ + P}}\left( {{\text{HTTH}}} \right) + {\text{P}}\left( {{\text{HTHT}}} \right) + {\text{P}}\left( {{\text{THTH}}} \right)\\&= \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} = \frac{6}{{16}} = \frac{3}{8}\\\\{\text{P}}\left( {{\text{X}} = 3} \right) &= {\text{P}}\left( {{\text{HHHT}}} \right) + {\text{P}}\left( {{\text{HHTH}}} \right) + {\text{P}}\left( {{\text{HTHH}}} \right) + {\text{P}}\left( {{\text{THHH}}} \right)\\& = \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} + \frac{1}{{16}} = \frac{4}{{16}} = \frac{1}{4}\\\\{\text{P}}\left( {{\text{X}} = 4} \right)& = {\text{P}}\left( {{\text{HHHH}}} \right) = \frac{1}{{16}}\end{align}

Therefore, the probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ P (X) $$\frac{1}{{16}}$$ $$\frac{1}{4}$$ $$\frac{3}{8}$$ $$\frac{1}{4}$$ $$\frac{1}{{16}}$$

## Chapter 13 Ex.13.4 Question 5

Find the probability distribution of the number of success in two tosses of die, where a success is defined as

(i) number greater than $$4$$

(ii) six appears on at least one die

### Solution

When a die is tossed two times, we obtain $$(6 × 6) = 36$$ number of observations.

Let X be the random variable, which represents the number of success.

(i) Here, success refers to the number greater than $$4$$.

$${\text{P}}\left( {{\text{X}} = 0} \right) =$$ P (number less than or equal to $$4$$ on both the tosses) =$$= \frac{4}{6} \times \frac{4}{6} = \frac{4}{9}$$

$${\text{P}}\left( {{\text{X}} = 1} \right) =$$P (number less than or equal to $$4$$ on first toss and greater than $$4$$ on second toss) + P (number greater than $$4$$ on first toss and less than or equal to 4 on second toss) $$= \frac{4}{6} \times \frac{2}{6} + \frac{4}{6} \times \frac{2}{6} = \frac{4}{9}$$

$${\text{P}}\left( {{\text{X}} = 2} \right) =$$P (number greater than $$4$$ on both the tosses)$$= \frac{2}{6} \times \frac{2}{6} = \frac{1}{9}$$

Therefore, the probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ P (X) $$\frac{4}{9}$$ $$\frac{4}{9}$$ $$\frac{1}{9}$$

(ii) Here, success means six appears on at least one die.

$${\text{P}}\left( {{\text{Y}} = 0} \right) =$$ P (six appears on none of the dice) $$= \frac{5}{6} \times \frac{5}{6} = \frac{{25}}{{36}}$$

$${\text{P}}\left( {{\text{Y}} = 1} \right) =$$ P (six appears on at least one of the dice)$$= \frac{1}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6} = \frac{5}{{36}} + \frac{5}{{36}} = \frac{{10}}{{36}}$$

Therefore, the required probability distribution is as follows.

 Y $$0$$ $$1$$ P (Y) $$\frac{{25}}{{36}}$$ $$\frac{{10}}{{36}}$$

## Chapter 13 Ex.13.4 Question 6

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

### Solution

Given, out of $$30$$ bulbs, $$6$$ are defective.

$$\therefore$$Number of non-defective bulbs$$= {\text{3}}0 - {\text{6}} = {\text{24}}$$

$$4$$ bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

\begin{align}{\text{P}}\left( {{\text{X}} = 0} \right) &= {\text{P}}\left( {{\text{4 non - defective and 0 defective}}} \right) = {}^4{{\text{C}}_0} \times {\left( {\frac{4}{5}} \right)^4} = \frac{{256}}{{625}}\\{\text{P}}\left( {{\text{X}} = 1} \right) &= {\text{P}}\left( {{\text{3 non - defective and 1 defective}}} \right) = {}^4{{\text{C}}_1} \times \left( {\frac{1}{5}} \right) \times {\left( {\frac{4}{5}} \right)^3} = \frac{{256}}{{625}}\\{\text{P}}\left( {{\text{X}} = 2} \right) &= {\text{P}}\left( {{\text{2 non - defective and 2 defective}}} \right) = {}^4{{\text{C}}_2} \times {\left( {\frac{1}{5}} \right)^2} \times {\left( {\frac{4}{5}} \right)^2} = \frac{{96}}{{625}}\\{\text{P}}\left( {{\text{X}} = 3} \right) &= {\text{P}}\left( {{\text{1 non - defective and 3 defective}}} \right) = {}^4{{\text{C}}_3} \times {\left( {\frac{1}{5}} \right)^3} \times \left( {\frac{4}{5}} \right) = \frac{{16}}{{625}}\\{\text{P}}\left( {{\text{X}} = 4} \right) &= {\text{P}}\left( {{\text{0 non - defective and 4 defective}}} \right) = {}^4{{\text{C}}_4} \times {\left( {\frac{1}{5}} \right)^4} = \frac{1}{{625}}\end{align}

Thus, the required probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ P (X) $$\frac{{256}}{{625}}$$ $$\frac{{256}}{{625}}$$ $$\frac{{96}}{{625}}$$ $$\frac{{16}}{{625}}$$ $$\frac{1}{{625}}$$

## Chapter 13 Ex.13.4 Question 7

A coin is biased so that the head is $$3$$ times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

### Solution

Let the probability of getting a tail in the biased coin be x.

\begin{align}\therefore {\text{P}}\left( {\text{T}} \right) &= x\\ \Rightarrow {\text{P}}\left( {\text{H}} \right) &= 3x\end{align}

For a biased coin, $${\text{P}}\left( {\text{T}} \right) + {\text{P}}\left( {\text{H}} \right) = 1$$

\begin{align}& \Rightarrow \;x + 3x = 1\\ &\Rightarrow \;4x = 1\\& \Rightarrow \;x = \frac{1}{4}\\&\therefore {\text{P}}\left( {\text{T}} \right) = \frac{1}{4}\;{\text{and P}}\left( {\text{H}} \right) = \frac{3}{4}\end{align}

When the coin is tossed twice, the sample space is $$\left\{ {{\text{HH, TT, HT, TH}}} \right\}$$.

Let X be the random variable representing the number of tails.

\begin{align}{\text{P}}\left( {{\text{X}} = 0} \right) &= {\text{P}}\left( {{\text{no tail}}} \right) = {\text{P}}\left( {\text{H}} \right) \times {\text{P}}\left( {\text{H}} \right) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{{16}}\\{\text{P}}\left( {{\text{X}} = 1} \right)& = {\text{P}}\left( {{\text{one}}\;{\text{tail}}} \right) = {\text{P}}\left( {{\text{HT}}} \right) + {\text{P}}\left( {{\text{TH}}} \right)\\&= \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}\\&= \frac{3}{{16}} + \frac{3}{{16}}\\&= \frac{3}{8}\\{\text{P}}\left( {{\text{X}} = 2} \right)& = {\text{P}}\left( {{\text{two tails}}} \right) = {\text{P}}\left( {{\text{TT}}} \right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{{16}}\end{align}

Thus, the required probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ P (X) $$\frac{9}{{16}}$$ $$\frac{3}{8}$$ $$\frac{1}{{16}}$$

## Chapter 13 Ex.13.4 Question 8

A random variable X has the following probability distribution.

 X $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ P (X) 0 $$k$$ $$2k$$ $$2k$$ $$3k$$ $${k^2}$$ $$2{k^2}$$ $$7{k^2} + k$$

Determine

(i) $$k$$

(ii) $${\text{P}}\left( {{\text{X}} < 3} \right)$$

(iii) $${\text{P}}\left( {{\text{X}} > {\text{6}}} \right)$$

(iv) $${\text{P}}\left( {0 < {\text{X}} < 3} \right)$$

### Solution

(i) Since, the sum of probabilities of a probability distribution of random variable is one.

\begin{align}&\therefore 0 + k + 2k + 3k + {k^2} + 2{k^2} + \left( {7{k^2} + k} \right) = 1\\& \Rightarrow \;10{k^2} + 9k - 1 = 0\\& \Rightarrow \;\left( {10k - 1} \right)\left( {k + 1} \right) = 0\\& \Rightarrow \;k = - 1,\;\frac{1}{{10}}\end{align}

(ii) $${\text{P}}\left( {{\text{X}} < 3} \right) = {\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right) + {\text{P}}\left( {{\text{X}} = 2} \right)$$

\begin{align} &= 0 + k + 2k\\ &= 3k\\ &= 3 \times \frac{1}{{10}}\\ &= \frac{3}{{10}}\end{align}

(iii) $${\text{P}}\left( {{\text{X}} > {\text{6}}} \right) = {\text{P}}\left( {{\text{X}} = 7} \right)$$

\begin{align}& = 7{k^2} + k\\ &= 7 \times {\left( {\frac{1}{{10}}} \right)^2} + \frac{1}{{10}}\\& = \frac{7}{{100}} + \frac{1}{{10}}\\ &= \frac{{17}}{{100}}\end{align}

(iv) $${\text{P}}\left( {0 < {\text{X}} < 3} \right) = {\text{P}}\left( {{\text{X}} = 1} \right) + {\text{P}}\left( {{\text{X}} = 2} \right)$$

\begin{align} &= k + 2k\\ = 3k\\ &= 3 \times \frac{1}{{10}}\\& = \frac{3}{{10}}\end{align}

## Chapter 13 Ex.13.4 Question 9

The random variable X has probability $${\text{P}}\left( {\text{X}} \right)$$ of the following form, where k is some number:

{\text{P}}\left( {\text{X}} \right) = \left\{ \begin{align}k,\;{\text{if}}\;x = 0\\2k,\;{\text{if}}\;x = 1\\3k,\;{\text{if}}\;x = 2\\0,\;{\text{otherwise}}\end{align} \right\}

(a) Determine the final value of k.

(b) Find $${\text{P}}\left( {{\text{X}} < 2} \right),\;{\text{P}}\left( {{\text{X}} \le 2} \right),\;{\text{P}}\left( {{\text{X}} \ge 2} \right)$$.

### Solution

(a) Since, the sum of probabilities of a probability distribution of random variable is one.

\begin{align}&\therefore k + 2k + 3k + 0 = 1\\& \Rightarrow \;6k = 1\\& \Rightarrow \;k = \frac{1}{6}\end{align}

(b) $${\text{P}}\left( {{\text{X}} < 2} \right) = {\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)$$

\begin{align}&\therefore k + 2k\\& \Rightarrow \;3k = \frac{3}{6} = \frac{1}{2}\end{align}

\begin{align}{\text{P}}\left( {{\text{X}} \le 2} \right)& = {\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right) + {\text{P}}\left( {{\text{X}} = 2} \right)\\&\therefore k + 2k + 3k\\& \Rightarrow \;6k = \frac{6}{6} = 1\end{align}

\begin{align}{\text{P}}\left( {{\text{X}} \ge 2} \right)& = {\text{P}}\left( {{\text{X}} = 2} \right) + {\text{P}}\left( {{\text{X}} > 2} \right)\\& = 3k + 0\\ &\Rightarrow \;3k = \frac{3}{6} = \frac{1}{2}\end{align}

## Chapter 13 Ex.13.4 Question 10

Find the mean number of heads in three tosses of a fair coin.

### Solution

Let X denote the success of getting heads.

Thus, the sample space is $${\text{S}} = \left\{ {{\text{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}}} \right\}$$

Here, X can take the value of $$0, 1, 2,$$ or $$3$$.

\begin{align}&\therefore {\text{P}}\left( {{\text{X}} = 0} \right) = {\text{P}}\left( {{\text{TTT}}} \right)\\ &= {\text{P}}\left( {\text{T}} \right) \times {\text{P}}\left( {\text{T}} \right) \times {\text{P}}\left( {\text{T}} \right)\\ &= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\end{align}

\begin{align}\therefore {\text{P}}\left( {{\text{X}} = 1} \right)& = {\text{P}}\left( {{\text{HTT}}} \right) + {\text{P}}\left( {{\text{THT}}} \right) + {\text{P}}\left( {{\text{TTH}}} \right)\\ &= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\\ &= \frac{3}{8}\end{align}

\begin{align}\therefore {\text{P}}\left( {{\text{X}} = 2} \right)& = {\text{P}}\left( {{\text{HHT}}} \right) + {\text{P}}\left( {{\text{HTH}}} \right) + {\text{P}}\left( {{\text{THH}}} \right)\\ &= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\\ &= \frac{3}{8}\end{align}

\begin{align}\therefore {\text{P}}\left( {{\text{X}} = 3} \right) &= {\text{P}}\left( {{\text{HHH}}} \right)\\& = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\\ &= \frac{1}{8}\end{align}

Thus, the required probability is as follows.

 X $$0$$ $$1$$ $$2$$ $$3$$ P (X) $$\frac{1}{8}$$ $$\frac{3}{8}$$ $$\frac{3}{8}$$ $$\frac{1}{8}$$

\begin{align}{\text{Mean E}}\left( {\text{X}} \right) &= \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)\\ &= 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8}\\& = \frac{3}{8} + \frac{3}{4} + \frac{3}{8}\\ &= \frac{{12}}{8} = \frac{3}{2} = 1.5\end{align}

## Chapter 13 Ex.13.4 Question 11

Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

### Solution

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Thus, X can take the value of $$0$$, $$1$$ or $$2$$.

\begin{align}\therefore {\text{P}}\left( {{\text{X}} = 0} \right) &= {\text{P}}\left( {{\text{not getting six on any of the dice}}} \right) = \frac{{25}}{{36}}\\{\text{P}}\left( {{\text{X}} = 1} \right) &= {\text{P}}\left( {{\text{six on first die and no six on second die}}} \right) + {\text{P}}\left( {{\text{no six on first die and six on second die}}} \right)\\&= 2 \times \left( {\frac{1}{6} \times \frac{5}{6}} \right) = \frac{{10}}{{36}}\\{\text{P}}\left( {{\text{X}} = 2} \right)& = {\text{P}}\left( {{\text{six on both the dice}}} \right) = \frac{1}{{36}}\end{align}

Thus, the required probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ P (X) $$\frac{{25}}{{36}}$$ $$\frac{{10}}{{36}}$$ $$\frac{1}{{36}}$$

Then, expectation of $${\text{X}} = {\text{E}}\left( {\text{X}} \right) = \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)$$

\begin{align}& = 0 \times \frac{{25}}{{36}} + 1 \times \frac{{10}}{{36}} + 2 \times \frac{1}{{36}}\\& = \frac{1}{3}\end{align}

## Chapter 13 Ex.13.4 Question 12

Two numbers are selected at random (without replacement) from the first six positive integers. Let $$X$$ denotes the larger of two numbers obtained. Find  $${\text{E}}\left( {\text{X}} \right)$$.

### Solution

The two positive integers can be selected from the first six positive integers without replacement in $$6 \times 5 = 30\;{\text{ways}}$$.

$$X$$ represents the larger of the two numbers obtained.

Thus, $$X$$ can take the value of $$2, 3, 4, 5$$ or $$6$$.

For $${\text{X}} = 2$$, the possible observations are$$\left( {1,2} \right)\;{\text{and}}\;\left( {2,1} \right)$$.

$$\therefore {\text{P}}\left( {{\text{X}} = 2} \right) = \frac{2}{{30}} = \frac{1}{{15}}$$

For $${\text{X}} = 3$$, the possible observations are$$\left( {1,3} \right),\;\left( {2,3} \right),\;\left( {3,1} \right)\;{\text{and}}\left( {3,2} \right)$$.

$$\therefore {\text{P}}\left( {{\text{X}} = 3} \right) = \frac{4}{{30}} = \frac{2}{{15}}$$

For $${\text{X}} = 4$$, the possible observations are$$\left( {1,4} \right),\;\left( {2,4} \right),\;\left( {3,4} \right),\;\left( {4,3} \right),\;\left( {4,2} \right)\;{\text{and}}\left( {4,1} \right)$$.

$$\therefore {\text{P}}\left( {{\text{X}} = 4} \right) = \frac{6}{{30}} = \frac{1}{5}$$

For $${\text{X}} = 5$$, the possible observations are$$\left( {1,5} \right),\;\left( {2,5} \right),\;\left( {3,5} \right),\;\left( {4,5} \right),\;\left( {5,4} \right),\;\left( {5,3} \right),\;\left( {5,2} \right)\;{\text{and }}\left( {5,1} \right)$$

$$\therefore {\text{P}}\left( {{\text{X}} = 5} \right) = \frac{8}{{30}} = \frac{4}{{15}}$$

For $${\text{X}} = 6$$, the possible observations are $$\left( {1,6} \right),\;\left( {2,6} \right),\;\left( {3,6} \right),\;\left( {4,6} \right),\;\left( {5,6} \right),\;\left( {6,4} \right),\;\left( {6,3} \right),\;\left( {6,2} \right),\;\left( {6,1} \right)$$.

$$\therefore {\rm{P}}\left( {{\rm{X}} = 6} \right) = \frac{{10}}{{30}} = \frac{1}{3}$$

Thus, the required probability distribution is as follows.

 X $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ P (X) $$\frac{1}{{15}}$$ $$\frac{2}{{15}}$$ $$\frac{1}{5}$$ $$\frac{4}{{15}}$$ $$\frac{1}{3}$$

Then, $${\text{E}}\left( {\text{X}} \right) = \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)$$

\begin{align}& = 2 \times \frac{1}{{15}} + 3 \times \frac{2}{{15}} + 4 \times \frac{1}{5} + 5 \times \frac{4}{{15}} + 6 \times \frac{1}{3}\\ &= \frac{2}{{15}} + \frac{2}{5} + \frac{4}{5} + \frac{4}{3} + 2\\ &= \frac{{70}}{{15}} = \frac{{14}}{3}\end{align}

## Chapter 13 Ex.13.4 Question 13

Let X denotes the sum of the number obtained when two fair dice are rolled. Find the variance and standard deviation of X.

### Solution

Here, X can take values $$2, 3, 4, 5, 6, 7, 8, 9, 10, 11$$ and $$12$$.

\begin{align}{\text{P}}\left( {{\text{X}} = 2} \right) &= {\text{P}}\left( {1,1} \right) = \frac{1}{{36}}\\{\text{P}}\left( {{\text{X}} = 3} \right) &= {\text{P}}\left( {1,2} \right) + {\text{P}}\left( {2,1} \right) = \frac{2}{{36}} = \frac{1}{{18}}\\{\text{P}}\left( {{\text{X}} = 4} \right)& = {\text{P}}\left( {1,3} \right) + {\text{P}}\left( {2,2} \right) + {\text{P}}\left( {3,1} \right) = \frac{3}{{36}} = \frac{1}{{12}}\\{\text{P}}\left( {{\text{X}} = 5} \right)& = {\text{P}}\left( {1,4} \right) + {\text{P}}\left( {2,3} \right) + {\text{P}}\left( {3,2} \right) + {\text{P}}\left( {4,1} \right) = \frac{4}{{36}} = \frac{1}{9}\\{\text{P}}\left( {{\text{X}} = 6} \right) &= {\text{P}}\left( {1,5} \right) + {\text{P}}\left( {2,4} \right) + {\text{P}}\left( {3,3} \right) + {\text{P}}\left( {4,2} \right) + {\text{P}}\left( {5,1} \right) = \frac{5}{{36}}\\{\text{P}}\left( {{\text{X}} = 7} \right) &= {\text{P}}\left( {1,6} \right) + {\text{P}}\left( {2,5} \right) + {\text{P}}\left( {3,4} \right) + {\text{P}}\left( {4,3} \right) + {\text{P}}\left( {5,2} \right) + {\text{P}}\left( {6,1} \right) = \frac{6}{{36}} = \frac{1}{6}\\{\text{P}}\left( {{\text{X}} = 8} \right) &= {\text{P}}\left( {2,6} \right) + {\text{P}}\left( {3,5} \right) + {\text{P}}\left( {4,4} \right) + {\text{P}}\left( {5,3} \right) + {\text{P}}\left( {6,2} \right) = \frac{5}{{36}}\\{\text{P}}\left( {{\text{X}} = 9} \right) &= {\text{P}}\left( {3,6} \right) + {\text{P}}\left( {4,5} \right) + {\text{P}}\left( {5,4} \right) + {\text{P}}\left( {6,3} \right) = \frac{4}{{36}} = \frac{1}{9}\\{\text{P}}\left( {{\text{X}} = 10} \right) &= {\text{P}}\left( {4,6} \right) + {\text{P}}\left( {5,5} \right) + {\text{P}}\left( {6,4} \right) = \frac{3}{{36}} = \frac{1}{{12}}\\{\text{P}}\left( {{\text{X}} = 11} \right) &= {\text{P}}\left( {5,6} \right) + {\text{P}}\left( {6,5} \right) = \frac{2}{{36}} = \frac{1}{{18}}\\{\text{P}}\left( {{\text{X}} = 12} \right) &= {\text{P}}\left( {6,6} \right) = \frac{1}{{36}}\end{align}

Thus, the required probability distribution is as follows.

 X $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ P (X) $$\frac{1}{{36}}$$ $$\frac{1}{{18}}$$ $$\frac{1}{{12}}$$ $$\frac{1}{9}$$ $$\frac{5}{{36}}$$ $$\frac{1}{6}$$ $$\frac{5}{{36}}$$ $$\frac{1}{9}$$ $$\frac{1}{{12}}$$ $$\frac{1}{{18}}$$ $$\frac{1}{{36}}$$

Then, $${\text{E}}\left( {\text{X}} \right) = \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)$$

\begin{align}{\text{E}}\left( {\text{X}} \right)& = \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)\\ &= 2 \times \frac{1}{{36}} + 3 \times \frac{1}{{18}} + 4 \times \frac{1}{{12}} + 5 \times \frac{1}{9} + 6 \times \frac{5}{{36}} + 7 \times \frac{1}{6} \\& \qquad+ 8 \times \frac{5}{{36}} + 9 \times \frac{1}{9} + 10 \times \frac{1}{{12}} + 11 \times \frac{1}{{18}} + 12 \times \frac{1}{{36}}\\[5pt] &= \frac{1}{{18}} + \frac{1}{6} + \frac{1}{3} + \frac{5}{9} + \frac{5}{6} + \frac{7}{6} + \frac{{10}}{9} + 1 + \frac{5}{6} + \frac{{11}}{{18}} + \frac{1}{3}\\[5pt] &= 7\end{align}

\begin{align}{\text{E}}\left( {{{\text{X}}^2}} \right) &= \Sigma {{\text{X}}^2}_{\text{i}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)\\& = 4 \times \frac{1}{{36}} + 9 \times \frac{1}{{18}} + 16 \times \frac{1}{{12}} + 25 \times \frac{1}{9} + 36 \times \frac{5}{{36}} + 49 \times \frac{1}{6} + 64 \times \frac{5}{{36}}\\&\qquad + 81 \times \frac{1}{9} + 100 \times \frac{1}{{12}} + 121 \times \frac{1}{{18}} + 144 \times \frac{1}{{36}}\\[5pt] &= \frac{1}{9} + \frac{1}{2} + \frac{4}{3} + \frac{{25}}{9} + 5 + \frac{{49}}{6} + \frac{{80}}{9} + 9 + \frac{{25}}{3} + \frac{{121}}{{18}} + 4\\[5pt]& = \frac{{987}}{{18}} = \frac{{329}}{6} = 54.833\end{align}

Then, $${\text{Var}}\left( {\text{X}} \right) = {\text{E}}\left( {{{\text{X}}^2}} \right) - {\left[ {{\text{E}}\left( {\text{X}} \right)} \right]^2}$$

\begin{align} &= 54.833 - {7^2}\\&= 54.833 - 49\\ &= 5.833\end{align}

\begin{align}\therefore {\text{Standard}}\;{\text{Variation}}& = \sqrt {{\text{Var}}\left( {\text{X}} \right)} \\& = \sqrt {5.833} \\&= 2.415\end{align}

## Chapter 13 Ex.13.4 Question 14

A class has $$15$$ students whose ages are $$14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19$$ and $$20$$ years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

### Solution

There are $$15$$ students in the class. Each student has the same chance to be chosen.

Thus, the probability of each student to be selected is $$\frac{1}{{15}}$$.

The given information can be shown in the frequency table as follows.

 X $$14$$ $$15$$ $$16$$ $$17$$ $$18$$ $$19$$ $$20$$ $$21$$ f $$2$$ $$1$$ $$2$$ $$3$$ $$1$$ $$2$$ $$3$$ $$1$$

\begin{align}{\text{P}}\left( {{\text{X}} = 14} \right)& = \frac{2}{{15}}{\text{,}}\;{\text{P}}\left( {{\text{X}} = 15} \right) = \frac{1}{{15}},\;{\text{P}}\left( {{\text{X}} = 16} \right) = \frac{2}{{15}},\;{\text{P}}\left( {{\text{X}} = 17} \right) = \frac{3}{{15}}\\{\text{P}}\left( {{\text{X}} = 18} \right) &= \frac{1}{{15}}{\text{,}}\;{\text{P}}\left( {{\text{X}} = 19} \right) = \frac{2}{{15}},\;{\text{P}}\left( {{\text{X}} = 20} \right) = \frac{3}{{15}},\;{\text{P}}\left( {{\text{X}} = 21} \right) = \frac{1}{{15}}\end{align}

Thus, the probability distribution of random variable X is as follows.

 X $$14$$ $$15$$ $$16$$ $$17$$ $$18$$ $$19$$ $$20$$ $$21$$ P (X) $$\frac{2}{{15}}$$ $$\frac{1}{{15}}$$ $$\frac{2}{{15}}$$ $$\frac{3}{{15}}$$ $$\frac{1}{{15}}$$ $$\frac{2}{{15}}$$ $$\frac{3}{{15}}$$ $$\frac{1}{{15}}$$

Then,$${\text{Mean X}} = {\text{E}}\left( {\text{X}} \right)$$

\begin{align}& = \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)\\& = 14 \times \frac{2}{{15}} + 15 \times \frac{1}{{15}} + 16 \times \frac{2}{{15}} + 17 \times \frac{3}{{15}} + 18 \times \frac{1}{{15}} + 19 \times \frac{2}{{15}} + 20 \times \frac{3}{{15}} + 21 \times \frac{1}{{15}}\\ &= \frac{1}{{15}}\left( {28 + 15 + 32 + 51 + 18 + 38 + 60 + 21} \right)\\ &= \frac{{263}}{{15}} = 17.53\end{align}

\begin{align}{\text{E}}\left( {{{\text{X}}^2}} \right) &= \Sigma {{\text{X}}^2}_{\text{i}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)\\& = {\left( {14} \right)^2} \times \frac{2}{{15}} + {\left( {15} \right)^2} \times \frac{1}{{15}} + {\left( {16} \right)^2} \times \frac{2}{{15}} + {\left( {17} \right)^2} \times \frac{3}{{15}}\\& \qquad + {\left( {18} \right)^2} \times \frac{1}{{15}} + {\left( {19} \right)^2} \times \frac{2}{{15}} + {\left( {20} \right)^2} \times \frac{3}{{15}} + {\left( {21} \right)^2} \times \frac{1}{{15}}\\[5pt] &= \frac{1}{{15}}\left( {392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441} \right)\\[5pt] &= \frac{{4683}}{{15}} = 312.2\end{align}

\begin{align}\therefore {\text{Variance}}\left( {\text{X}} \right) &= {\text{E}}\left( {{{\text{X}}^2}} \right) - {\left[ {{\text{E}}\left( {\text{X}} \right)} \right]^2}\\&= 312.2 - {\left( {\frac{{263}}{{15}}} \right)^2}\\& = 312.2 - 307.4177\\&= 4.7823\\&= 4.78\end{align}

\begin{align}\therefore {\text{Standard}}\;{\text{Variation}} &= \sqrt {{\text{Variance}}\left( {\text{X}} \right)} \\&= \sqrt {4.78} \\&= 2.186 \approx 2.19\end{align}

## Chapter 13 Ex.13.4 Question 15

In a meeting, $$70\%$$ of the members favour and $$30\%$$ oppose a certain proposal. A member is selected at random and we take $${\text{X}} = 0$$if he opposed, and $${\text{X}} = 1$$if he is in favour. Find $${\text{E}}\left( {\text{X}} \right)$$and $${\mathop{\text var}} \left( {\text{X}} \right)$$.

### Solution

Given, $${\text{P}}\left( {{\text{X}} = 0} \right) = 30\% = \frac{{30}}{{100}} = 0.3$$

$${\text{P}}\left( {{\text{X}} = 1} \right) = 70\% = \frac{{70}}{{100}} = 0.7$$

Thus, the probability distribution is as follows.

 X $$0$$ $$1$$ P (X) $$0.3$$ $$0.7$$

$${\text{E}}\left( {\text{X}} \right) = \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)$$

\begin{align} &= 0 \times 0.3 + 1 \times 0.7\\& = 0.7\end{align}

\begin{align}{\text{E}}\left( {{{\text{X}}^2}} \right) &= \Sigma {{\text{X}}^2}_{\text{i}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)\\&= {0^2} \times 0.3 + {1^2} \times 0.7\\& = 0.7\end{align}

\begin{align}{\text{Var}}\left( {\text{X}} \right) &= {\text{E}}\left( {{{\text{X}}^2}} \right) - {\left[ {{\text{E}}\left( {\text{X}} \right)} \right]^2}\\& = 0.7 - {\left( {0.7} \right)^2}\\&= 0.7 - 0.49\\& = 0.21\end{align}

## Chapter 13 Ex.13.4 Question 16

The mean of the numbers obtained on throwing a die having written $$1$$ on three faces, $$2$$ on two faces and $$5$$ on one face is

(A) $$1$$

(B) $$2$$

(C) $$5$$

(D) $$\frac{8}{3}$$

### Solution

Let X be the random variable representing a number on the die.

The total number of observations is six.

$$\therefore {\text{P}}\left( {{\text{X}} = 1} \right) = \frac{3}{6} = \frac{1}{2}$$

$${\text{P}}\left( {{\text{X}} = 2} \right) = \frac{2}{6} = \frac{1}{3}$$

$${\text{P}}\left( {{\text{X}} = 5} \right) = \frac{1}{6}$$

Thus, the probability distribution is as follows.

 X $$1$$ $$2$$ $$5$$ P (X) $$\frac{1}{2}$$ $$\frac{1}{3}$$ $$\frac{1}{6}$$

$${\text{Mean}} = {\text{E}}\left( {\text{X}} \right) = \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)$$

\begin{align}& = \frac{1}{2} \times 1 + \frac{1}{3} \times 2 + \frac{1}{6} \times 5\\ &= \frac{1}{2} + \frac{2}{3} + \frac{5}{6}\\& = \frac{{12}}{6} = 2\end{align}

The correct answer is B.

## Chapter 13 Ex.13.4 Question 17

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of $${\text{E}}\left( {\text{X}} \right)$$is

(A) $$\frac{{37}}{{221}}$$

(B) $$\frac{5}{{13}}$$

(C) $$\frac{1}{{13}}$$

(D) $$\frac{2}{{13}}$$

### Solution

Let X denote the number of aces obtained.

Thus, X can take any of the values of $$0, 1$$, or $$2$$.

Since, in a deck of $$52$$ cards, $$4$$ cards are aces. Thus, there are $$48$$ non-ace cards.

\begin{align}\therefore {\text{P}}\left( {{\text{X}} = 0} \right) &= {\text{P}}\left( {{\text{0 ace and 2 non - ace cards}}} \right) = \frac{{{}^4{{\text{C}}_0} \times {}^{48}{{\text{C}}_2}}}{{{}^{52}{{\text{C}}_2}}} = \frac{{1128}}{{1326}}\\{\text{P}}\left( {{\text{X}} = 1} \right) &= {\text{P}}\left( {{\text{1 ace and 1 non - ace cards}}} \right) = \frac{{{}^4{{\text{C}}_1} \times {}^{48}{{\text{C}}_1}}}{{{}^{52}{{\text{C}}_2}}} = \frac{{192}}{{1326}}\\{\text{P}}\left( {{\text{X}} = 2} \right) &= {\text{P}}\left( {{\text{2 ace and 0 non - ace cards}}} \right) = \frac{{{}^4{{\text{C}}_2} \times {}^{48}{{\text{C}}_0}}}{{{}^{52}{{\text{C}}_2}}} = \frac{6}{{1326}}\end{align}

Thus, the probability distribution is as follows.

 X $$0$$ $$1$$ $$2$$ P (X) $$\frac{{1128}}{{1326}}$$ $$\frac{{192}}{{1326}}$$ $$\frac{6}{{1326}}$$

\begin{align}{\text{E}}\left( {\text{X}} \right) &= \Sigma {{\text{X}}_{\text{i}}}{\text{P}}\left( {{{\text{X}}_{\text{i}}}} \right)\\ &= 0 \times \frac{{1128}}{{1326}} + 1 \times \frac{{192}}{{1326}} + 2 \times \frac{6}{{1326}}\\ & = \frac{{204}}{{1326}} = \frac{2}{{13}}\end{align}

Thus, the correct answer is D.

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