Differentiation (Part-38) | S N De

Random Variable and Its Distribution (Class 12): MCQ, VSA and Short Answer Questions

Complete solutions to 6 MCQs (1 mark), 16 Very Short Answer questions (2 marks) and 2 Short Answer questions (4 marks) from the chapter Random Variable and Its Distribution. They cover the definition of a random variable, probability distribution, mean, variance, standard deviation and the key proofs \(E(ax+b)=aE(x)+b\), \(\mathrm{Var}(a+bx)=b^2\mathrm{Var}(x)\) and \(\mathrm{Var}(x)=E(x^2)-\{E(x)\}^2\). Useful for CBSE Class 12 board exam, JEE and other competitive exam preparation.

MCQ Q1–6VSA Q1–16Short Answer Q1–2

Part A: Multiple Choice Questions (Q1–Q6)

MCQ Q11 Mark

The probability function of a random variable is always —

• (A) \(\ge1\)
• (B) \(0>1\)
• (C) positive
• (D) non-negative

Solution

The probability function of a random variable \(X\) is \(f(x_i)=P(X=x_i)\). Every probability lies between 0 and 1, so \(f(x_i)\ge0\), and \(\sum f(x_i)=1\).

• (A) is wrong: \(f(x_i)\ge1\) is not true in general, since the values add up to 1.
• (B) is impossible.
• (C) is wrong: \(f(x_i)\) can be 0 for some value.
• (D) is correct: \(f(x_i)\ge0\) always.

Answer: (D) non-negative.

MCQ Q21 Mark

The value of \(E(10x)\) is —

• (A) \(10+E(x)\)
• (B) \(10E(x)\)
• (C) \(E(x)\)
• (D) \(E(x)-10\)

Solution

For constants \(a\) and \(b\), \(E(ax+b)=aE(x)+b\). Taking \(a=10,\ b=0\):

\[E(10x)=10E(x).\]

Answer: (B) \(10E(x)\).

MCQ Q31 Mark

The value of \(\mathrm{Var}(4x)\) is —

• (A) \(16\,\mathrm{Var}(x)\)
• (B) \(4\,\mathrm{Var}(x)\)
• (C) \(2\,\mathrm{Var}(x)\)
• (D) \(\mathrm{Var}(x)+4\)

Solution

For constants \(a\) and \(b\), \(\mathrm{Var}(a+bx)=b^2\,\mathrm{Var}(x)\). Taking \(a=0,\ b=4\):

\[\mathrm{Var}(4x)=4^2\,\mathrm{Var}(x)=16\,\mathrm{Var}(x).\]

Answer: (A) \(16\,\mathrm{Var}(x)\).

MCQ Q41 Mark

The value of \(\mathrm{Var}(x)\) is —

• (A) \(E(x^2)-E(x)\)
• (B) \(E(x^2)-\{E(x)\}^2\)
• (C) \(E(x^2)+E(x)\)
• (D) \(\{E(x)\}^2-E(x^2)\)

Solution

By definition, \(\mathrm{Var}(x)=E[(x-\mu)^2]\) with \(\mu=E(x)\). Expanding (proved in VSA Q14),

\[\mathrm{Var}(x)=E(x^2)-\{E(x)\}^2.\]

Option (D) is the negative of this, and the variance is never negative, so (D) cannot be right.

Answer: (B) \(E(x^2)-\{E(x)\}^2\). (In the book the option is printed as \(\{E(x)^2\}\), which is read as \(\{E(x)\}^2\).)

MCQ Q51 Mark

If \(f(x)\) is the probability function of a random variable \(X\) and \(X\) can assume only two values \(x_1, x_2\), then the value of \(f(x_1)+f(x_2)\) is —

• (A) \(\ge1\)
• (B) \(\le1\)
• (C) \(\ge0\)
• (D) \(1\)

Solution

The values of \(X\) are only \(x_1\) and \(x_2\), so these two cases cover the whole sample space. The total probability is 1:

\[f(x_1)+f(x_2)=P(X=x_1)+P(X=x_2)=1.\]

Answer: (D) \(1\).

MCQ Q61 Mark

\(\mathrm{Var}(ax+b)\) is equal to —

• (A) \(a\,\mathrm{Var}(x)+b\)
• (B) \(a^2\,\mathrm{Var}(x)+b\)
• (C) \(a^2\,\mathrm{Var}(x)\)
• (D) none of these

Solution

Adding the constant \(b\) shifts every value but does not change the spread, and multiplying by \(a\) multiplies the variance by \(a^2\). (Proof in VSA Q9.)

\[\mathrm{Var}(ax+b)=a^2\,\mathrm{Var}(x).\]

Answer: (C) \(a^2\,\mathrm{Var}(x)\).

Part B: Very Short Answer Type Questions (Q1–Q16)

VSA Q12 Marks

Write a short note on: Random Variable.

Solution

Definition

Let \(S\) be the sample space of a random experiment. A random variable \(X\) is a real-valued function \(X:S\to\mathbb{R}\) that assigns a real number \(X(\omega)\) to every outcome \(\omega\in S\).

Key points

• A random variable is a function, not an ordinary variable. It is usually written \(X, Y, Z\).
• It is discrete if it takes finitely many or countably many values (for example the number of heads), and continuous if it can take any value in an interval (for example height or time).
• The event "\(X=x\)" means the set of outcomes \(\{\omega: X(\omega)=x\}\).

Example

Toss two coins. \(S=\{HH,HT,TH,TT\}\). Let \(X\) be the number of heads. Then \(X(HH)=2,\ X(HT)=X(TH)=1,\ X(TT)=0\), so \(X\) takes the values \(0,1,2\).

A random variable assigns a real number to each outcome of a random experiment.

VSA Q22 Marks

Write a short note on: Probability Distribution of Random Variable.

Solution

Definition

Let a discrete random variable \(X\) take the values \(x_1,x_2,\dots,x_n\) with probabilities \(p_i=P(X=x_i)\). The list of values together with their probabilities is the probability distribution of \(X\).

\(X\)	\(x_1\)	\(x_2\)	\(\cdots\)	\(x_n\)
\(P(X)\)	\(p_1\)	\(p_2\)	\(\cdots\)	\(p_n\)

It must satisfy \(p_i\ge0\) for every \(i\) and \(\sum_{i=1}^{n}p_i=1\).

Example

For the number of heads \(X\) when two fair coins are tossed:

\(X\)	\(0\)	\(1\)	\(2\)
\(P(X)\)	\(\frac14\)	\(\frac12\)	\(\frac14\)

A probability distribution lists every value of \(X\) with its probability, and the probabilities add up to 1.

VSA Q32 Marks

Write a short note on: Mean or Mathematical Expectation of Random Variable.

Solution

Definition

If \(X\) takes the values \(x_1,\dots,x_n\) with probabilities \(p_1,\dots,p_n\), its mean or mathematical expectation is

\[E(X)=\mu=\sum_{i=1}^{n}x_ip_i=x_1p_1+x_2p_2+\cdots+x_np_n.\]

It is the weighted average of the values of \(X\), with the probabilities as weights. It shows the long-run average value of \(X\) over many repetitions of the experiment.

Properties

\(E(K)=K\) for a constant \(K\), and \(E(aX+b)=aE(X)+b\).

Example

For the number of heads in two tosses, \(E(X)=0\cdot\frac14+1\cdot\frac12+2\cdot\frac14=1\).

\(E(X)=\sum x_ip_i\).

VSA Q42 Marks

Write a short note on: Variance of Random Variable.

Solution

Definition

The variance of \(X\) measures how widely its values are spread around the mean \(\mu=E(X)\):

\[\mathrm{Var}(X)=E[(X-\mu)^2]=\sum_{i=1}^{n}(x_i-\mu)^2p_i=E(X^2)-\{E(X)\}^2.\]

The standard deviation is \(\sigma=\sqrt{\mathrm{Var}(X)}\). The variance is never negative.

Properties

\(\mathrm{Var}(K)=0\) for a constant \(K\), and \(\mathrm{Var}(a+bX)=b^2\,\mathrm{Var}(X)\).

Example

For the number of heads in two tosses, \(E(X^2)=0+1\cdot\frac12+4\cdot\frac14=\frac32\), so \(\mathrm{Var}(X)=\frac32-1^2=\frac12\).

\(\mathrm{Var}(X)=E(X^2)-\{E(X)\}^2\).

VSA Q52 Marks

Prove that \(E(K)=K\), where \(K\) is a constant.

Solution

A constant \(K\) is a random variable that takes the single value \(K\) with probability \(1\). Therefore

\[E(K)=K\times1=K.\]

(Equivalently, for any random variable with \(\sum p_i=1\): \(E(K)=\sum Kp_i=K\sum p_i=K\).)

Hence proved: \(E(K)=K\).

VSA Q62 Marks

Prove that \(E(ax+b)=aE(x)+b\), where \(a\) and \(b\) are constants.

Solution

Let \(x\) take the values \(x_i\) with probabilities \(p_i\), where \(\sum p_i=1\). The variable \(ax+b\) takes the values \(ax_i+b\) with the same probabilities, so

\[\begin{aligned}E(ax+b)&=\sum(ax_i+b)p_i=a\sum x_ip_i+b\sum p_i\\&=aE(x)+b\times1=aE(x)+b.\end{aligned}\]

Hence proved: \(E(ax+b)=aE(x)+b\).

VSA Q72 Marks

Prove that \(E(x-\bar x)=0\), where \(\bar x\) is the mean of a random variable \(x\).

Solution

Here \(\bar x=\mu=E(x)\) is a constant. Using \(E(ax+b)=aE(x)+b\) with \(a=1,\ b=-\mu\):

\[E(x-\bar x)=E(x)-\bar x=\mu-\mu=0.\]

Directly: \(\sum(x_i-\mu)p_i=\sum x_ip_i-\mu\sum p_i=\mu-\mu=0\).

Hence proved: \(E(x-\bar x)=0\).

VSA Q82 Marks

Prove that \(\mathrm{Var}(K)=0\), where \(K\) is a constant.

Solution

For the constant random variable \(K\), \(E(K)=K\) (VSA Q5). By the definition of variance,

\[\mathrm{Var}(K)=E[(K-E(K))^2]=E[(K-K)^2]=E(0)=0.\]

Alternatively, \(\mathrm{Var}(K)=E(K^2)-\{E(K)\}^2=K^2-K^2=0\).

Hence proved: \(\mathrm{Var}(K)=0\).

VSA Q92 Marks

Prove that \(\mathrm{Var}(a+bx)=b^2\,\mathrm{Var}(x)\).

Solution

Let \(\mu=E(x)\). Then \(E(a+bx)=a+b\mu\). By the definition of variance,

\[\begin{aligned}\mathrm{Var}(a+bx)&=E\big[(a+bx)-(a+b\mu)\big]^2=E\big[b(x-\mu)\big]^2\\&=b^2E\big[(x-\mu)^2\big]=b^2\,\mathrm{Var}(x).\end{aligned}\]

Hence proved: \(\mathrm{Var}(a+bx)=b^2\,\mathrm{Var}(x)\).

VSA Q102 Marks

Define a random variable and explain what is meant by its mean (or, mathematical expectation).

Solution

Random variable

A random variable \(X\) is a real-valued function defined on the sample space \(S\) of a random experiment: it assigns a real number \(X(\omega)\) to each outcome \(\omega\in S\).

Mean (mathematical expectation)

If \(X\) takes the values \(x_1,\dots,x_n\) with probabilities \(p_1,\dots,p_n\), then

\[E(X)=\sum_{i=1}^{n}x_ip_i.\]

It is the average of the values of \(X\) when each value is weighted by its probability. If the experiment is repeated a very large number of times, the average of the observed values of \(X\) gets close to \(E(X)\).

Example

Let \(X\) be the number on a fair die. Each of \(1,\dots,6\) has probability \(\frac16\), so

\[E(X)=\frac{1+2+3+4+5+6}{6}=\frac72=3.5.\]

The mean \(E(X)=\sum x_ip_i\) is the probability-weighted average value of \(X\).

VSA Q112 Marks

Define a discrete random variable \(X\) and its mean \(E(X)\) and variance \(V(X)\).

Solution

Discrete random variable

A random variable is discrete if it takes only a finite number of values, or a countably infinite sequence of values, for example \(0,1,2,3,\dots\)

Mean\[E(X)=\mu=\sum_{i}x_ip_i,\quad\text{where } p_i=P(X=x_i).\]Variance\[V(X)=E[(X-\mu)^2]=\sum_i(x_i-\mu)^2p_i=E(X^2)-\{E(X)\}^2,\]

where \(E(X^2)=\sum x_i^2p_i\). Its positive square root is the standard deviation \(\sigma\).

\(E(X)=\sum x_ip_i\) and \(V(X)=\sum (x_i-\mu)^2p_i=E(X^2)-\{E(X)\}^2\).

VSA Q122 Marks

What is meant by discrete probability distribution?

Solution

If a random variable \(X\) can take only the discrete values \(x_1,x_2,\dots,x_n\), the rule that gives the probability \(p_i=P(X=x_i)\) of each value is called the discrete probability distribution of \(X\). It is usually shown as a table:

\(X\)	\(x_1\)	\(x_2\)	\(\cdots\)	\(x_n\)
\(P(X=x_i)\)	\(p_1\)	\(p_2\)	\(\cdots\)	\(p_n\)

with \(p_i\ge0\) and \(p_1+p_2+\cdots+p_n=1\).

Example

Three fair coins are tossed and \(X\) is the number of heads.

\(X\)	\(0\)	\(1\)	\(2\)	\(3\)
\(P(X)\)	\(\frac18\)	\(\frac38\)	\(\frac38\)	\(\frac18\)

A discrete probability distribution gives all values of \(X\) and their probabilities, which are non-negative and add up to 1.

VSA Q132 Marks

Define the probability function of a discrete random variable \(X\). State the conditions satisfied by it.

Solution

Definition

If \(X\) is a discrete random variable taking the values \(x_1,x_2,\dots,x_n\), the function \(f\) defined by

\[f(x_i)=P(X=x_i)=p_i,\qquad i=1,2,\dots,n,\]

is called the probability function (probability mass function) of \(X\).

Conditions

1. \(f(x_i)\ge0\) for every \(i\).
2. \(\sum_{i=1}^{n}f(x_i)=1\).

Example

\(f(x)=\dfrac{x}{6}\) for \(x=1,2,3\) is a probability function, because every value is positive and \(\frac16+\frac26+\frac36=1\).

The two conditions are \(f(x_i)\ge0\) and \(\sum f(x_i)=1\).

VSA Q142 Marks

The variance of a random variable \(x\) is defined by the expected value of \((x-\bar x)^2\), where \(\bar x\) is the mean of \(x\). Prove that \(\mathrm{Variance}(x)=E(x^2)-\{E(x)\}^2\).

Solution

Let \(\bar x=\mu=E(x)\), a constant. By definition,

\[\begin{aligned}\mathrm{Var}(x)&=E[(x-\mu)^2]=E[x^2-2\mu x+\mu^2]\\&=E(x^2)-2\mu E(x)+\mu^2\qquad(\text{since }\mu\text{ is a constant})\\&=E(x^2)-2\mu\cdot\mu+\mu^2\\&=E(x^2)-\mu^2.\end{aligned}\]

Since \(\mu=E(x)\),

Hence proved: \(\mathrm{Var}(x)=E(x^2)-\{E(x)\}^2\).

VSA Q152 Marks

Define the standard deviation of a random variable \(X\). Prove that \(\sigma=\sqrt{E(X^2)-\{E(X)\}^2}\), where the symbols have their usual meanings.

Solution

Definition

The standard deviation of \(X\), denoted \(\sigma\), is the positive square root of its variance:

\[\sigma=\sqrt{\mathrm{Var}(X)}=\sqrt{E[(X-\mu)^2]},\qquad \mu=E(X).\]Proof

We have \(\mathrm{Var}(X)=E[(X-\mu)^2]=E(X^2)-2\mu E(X)+\mu^2=E(X^2)-\mu^2\) (as proved in VSA Q14). Taking the positive square root,

\[\sigma=\sqrt{E(X^2)-\{E(X)\}^2}.\]Hence proved.

VSA Q162 Marks

If \(X\) and \(Y\) are independent random variables, show that \(\mathrm{Var}(aX+bY)=a^2\sigma_1^2+b^2\sigma_2^2\), where \(\sigma_1\) and \(\sigma_2\) are standard deviations of \(X\) and \(Y\) respectively and \(a, b\) are constants.

Solution

Let \(\mu_1=E(X)\) and \(\mu_2=E(Y)\), so \(\mathrm{Var}(X)=\sigma_1^2\) and \(\mathrm{Var}(Y)=\sigma_2^2\). Then \(E(aX+bY)=a\mu_1+b\mu_2\), and

\[\begin{aligned}\mathrm{Var}(aX+bY)&=E\big[(aX+bY)-(a\mu_1+b\mu_2)\big]^2=E\big[a(X-\mu_1)+b(Y-\mu_2)\big]^2\\&=a^2E(X-\mu_1)^2+b^2E(Y-\mu_2)^2+2ab\,E\big[(X-\mu_1)(Y-\mu_2)\big]\\&=a^2\sigma_1^2+b^2\sigma_2^2+2ab\,E\big[(X-\mu_1)(Y-\mu_2)\big].\end{aligned}\]

Since \(X\) and \(Y\) are independent, \(X-\mu_1\) and \(Y-\mu_2\) are independent, so the expectation of their product is the product of their expectations:

\[E\big[(X-\mu_1)(Y-\mu_2)\big]=E(X-\mu_1)\cdot E(Y-\mu_2)=0\times0=0\]

by VSA Q7. Therefore the last term vanishes.

Hence shown: \(\mathrm{Var}(aX+bY)=a^2\sigma_1^2+b^2\sigma_2^2\).

Part C: Short Answer Type Questions (Q1–Q2)

SA Q14 Marks

Define a random variable on the sample space of the random experiment of tossing two unbiased coins.

Solution

Step 1: Sample space

When two unbiased coins are tossed, the sample space is \(S=\{HH,\ HT,\ TH,\ TT\}\), and each outcome has probability \(\frac14\).

Step 2: Define the random variable

Let \(X\) be the number of heads obtained. Then \(X:S\to\mathbb{R}\) is given by

Outcome	\(HH\)	\(HT\)	\(TH\)	\(TT\)
\(X\) (number of heads)	\(2\)	\(1\)	\(1\)	\(0\)

So \(X\) can take the values \(0, 1, 2\), and each outcome of the experiment is assigned exactly one real number.

Step 3: Its probability distribution

\(P(X=0)=P(\{TT\})=\frac14\), \(P(X=1)=P(\{HT,TH\})=\frac24=\frac12\), \(P(X=2)=P(\{HH\})=\frac14\).

\(X\)	\(0\)	\(1\)	\(2\)
\(P(X)\)	\(\frac14\)	\(\frac12\)	\(\frac14\)

The probabilities add up to \(\frac14+\frac12+\frac14=1\). The mean is \(E(X)=0+\frac12+\frac12=1\).

Answer: \(X=\) number of heads is a random variable on \(S=\{HH,HT,TH,TT\}\) with \(X(HH)=2,\ X(HT)=X(TH)=1,\ X(TT)=0\).

SA Q24 Marks

Define two random variables \(X\) and \(Y\) on the sample space of the random experiment of rolling a balanced die.

Solution

Step 1: Sample space

When a balanced die is rolled, \(S=\{1,2,3,4,5,6\}\), and each outcome has probability \(\frac16\).

Step 2: Define X and Y

Many different random variables can be defined on the same sample space. Take

• \(X\): the number shown on the die, i.e. \(X(\omega)=\omega\).
• \(Y\): the indicator of an even number, i.e. \(Y(\omega)=1\) if \(\omega\) is even and \(Y(\omega)=0\) if \(\omega\) is odd.

Outcome \(\omega\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\(X(\omega)\)	1	2	3	4	5	6
\(Y(\omega)\)	0	1	0	1	0	1

Step 3: Their probability distributions

\(X\) takes each of the values \(1,\dots,6\) with probability \(\frac16\). \(Y\) takes the value \(1\) for \(\{2,4,6\}\) and \(0\) for \(\{1,3,5\}\).

\(X\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\(P(X)\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)

\(Y\)	\(0\)	\(1\)
\(P(Y)\)	\(\frac12\)	\(\frac12\)

For these, \(E(X)=\frac{21}{6}=\frac72\) and \(E(Y)=0\cdot\frac12+1\cdot\frac12=\frac12\).

Answer: On \(S=\{1,2,3,4,5,6\}\), \(X=\) number on the die and \(Y=1\) (even) or \(0\) (odd) are two random variables.

Formulas at a glance: \(E(X)=\sum x_ip_i\); \(E(aX+b)=aE(X)+b\); \(\mathrm{Var}(X)=E(X^2)-\{E(X)\}^2\); \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\); for independent \(X,Y\): \(\mathrm{Var}(aX+bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)\).