NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.4

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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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