Differentiation (Part-38) | S N De

Random Variable and Its Distribution (Class 12): Long Answer Type Questions 3–10

Step-by-step solutions to eight 5-mark long answer questions from the chapter Random Variable and Its Distribution: expectation and fair games, a dice-throwing prize problem, mean, variance and standard deviation of a distribution, transformed variables \(Y=aX+b\), independent variables and the expected value of a product. Each solution has the working and a boxed final answer. Useful for CBSE Class 12 board exam, JEE and other competitive exam preparation.

Q35 Marks

\(A\) and \(B\) play a game in which they take turns to throw 4 coins simultaneously. When \(A\) throws he receives ₹\(h\), where \(h\) is the number of heads shown. Find the expectation of \(A\). When \(B\) throws 4 heads or 4 tails he receives ₹\(k\); otherwise he receives nothing. Find the value of \(k\) for the game to be fair.

Solution

Step 1: Expectation of A

Let \(h\) be the number of heads when 4 fair coins are thrown. There are \(2^4=16\) equally likely outcomes, and \(P(h=r)=\dfrac{\binom4r}{16}\).

\(h\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(P(h)\)	\(\frac1{16}\)	\(\frac4{16}\)	\(\frac6{16}\)	\(\frac4{16}\)	\(\frac1{16}\)

\[E(A)=\frac{0\cdot1+1\cdot4+2\cdot6+3\cdot4+4\cdot1}{16}=\frac{32}{16}=2.\]

So \(A\) expects to receive ₹2 per throw.

Step 2: Expectation of B

\(B\) receives ₹\(k\) only for 4 heads (probability \(\frac1{16}\)) or 4 tails (probability \(\frac1{16}\)). So he receives ₹\(k\) with probability \(\frac1{16}+\frac1{16}=\frac18\), and ₹0 otherwise.

Amount	\(k\)	\(0\)
Probability	\(\frac18\)	\(\frac78\)

\[E(B)=k\cdot\frac18+0\cdot\frac78=\frac k8.\]Step 3: Condition for a fair game

The game is fair when the two players have equal expectations:

\[\frac k8=2\ \Rightarrow\ k=16.\]

Answer: \(E(A)=₹2\), and the game is fair when \(k=16\).

Q45 Marks

\(A\) and \(B\) play for a prize of ₹99. The prize is to be won by a player who first throws a "3" with one die. \(A\) first throws and if he fails, \(B\) throws and if he fails, \(A\) again throws and so on. Find their respective expectations.

Solution

Step 1: Probability that A wins

On each throw, \(P(3)=\frac16\) and \(P(\text{not }3)=\frac56\). \(A\) throws on turns 1, 3, 5, \(\dots\), so he wins on the 1st throw, or after two failures (his own and \(B\)'s), and so on:

\[P(A\text{ wins})=\frac16+\frac56\cdot\frac56\cdot\frac16+\left(\frac56\right)^4\frac16+\cdots\]

This is an infinite geometric series with first term \(\frac16\) and common ratio \(\frac{25}{36}\):

\[P(A\text{ wins})=\frac{1/6}{1-25/36}=\frac{1/6}{11/36}=\frac6{11}.\]Step 2: Probability that B wins

The game ends with a winner, so \(P(B\text{ wins})=1-\frac6{11}=\frac5{11}\). (Directly: \(\frac56\cdot\frac16+\left(\frac56\right)^3\frac16+\cdots=\dfrac{5/36}{11/36}=\dfrac5{11}\).)

Step 3: Expectations\[E(A)=99\times\frac6{11}=54,\qquad E(B)=99\times\frac5{11}=45.\]

Check: \(54+45=99\).

Answer: \(A\) expects ₹54 and \(B\) expects ₹45.

Q55 Marks

The probability distribution \(f(x)\) of a discrete random variable \(X\) is given by

\(x\)	1	2	3
\(p(x)\)	\(\frac12\)	\(\frac13\)	\(\frac16\)

Find the mean and variance of \(X\).

Solution

Check: \(\frac12+\frac13+\frac16=\frac{3+2+1}{6}=1\).

Mean\[E(X)=\sum xp(x)=1\cdot\frac12+2\cdot\frac13+3\cdot\frac16=\frac12+\frac23+\frac12=\frac53.\]Variance\[E(X^2)=\sum x^2p(x)=1\cdot\frac12+4\cdot\frac13+9\cdot\frac16=\frac12+\frac43+\frac32=\frac{10}3.\]\[\mathrm{Var}(X)=E(X^2)-\{E(X)\}^2=\frac{10}3-\frac{25}9=\frac{30-25}9=\frac59.\]

Answer: Mean \(=\dfrac53\), Variance \(=\dfrac59\).

Q65 MarksCBSE '07 (part ii)

(i) The probability distribution of a discrete random variable \(X\) is as follows:

\(x\)	\(-2\)	\(-1\)	0	2	4
\(p(x)\)	0.1	0.2	0.4	0.25	0.05

Find the expectation and S.D. of \(X\).
(ii) Find the mean, variance and standard deviation of the following probability distribution:

\(x\)	0	1	2	3
\(p(x)\)	\(\frac16\)	\(\frac12\)	\(\frac3{10}\)	\(\frac1{30}\)

Solution

(i)

Check: \(0.1+0.2+0.4+0.25+0.05=1\).

\[E(X)=(-2)(0.1)+(-1)(0.2)+0+2(0.25)+4(0.05)=-0.2-0.2+0.5+0.2=0.3.\]\[E(X^2)=4(0.1)+1(0.2)+0+4(0.25)+16(0.05)=0.4+0.2+1.0+0.8=2.4.\]\[\mathrm{Var}(X)=2.4-(0.3)^2=2.4-0.09=2.31,\qquad \text{S.D.}=\sqrt{2.31}\approx1.52.\]

Answer (i): \(E(X)=0.3\), S.D. \(=\sqrt{2.31}\approx1.52\).

(ii)

Check: \(\frac16+\frac12+\frac3{10}+\frac1{30}=\frac{5+15+9+1}{30}=1\).

\[E(X)=0+1\cdot\frac12+2\cdot\frac3{10}+3\cdot\frac1{30}=\frac12+\frac35+\frac1{10}=\frac{5+6+1}{10}=\frac65.\]\[E(X^2)=0+1\cdot\frac12+4\cdot\frac3{10}+9\cdot\frac1{30}=\frac12+\frac65+\frac3{10}=\frac{5+12+3}{10}=2.\]\[\mathrm{Var}(X)=2-\left(\frac65\right)^2=2-\frac{36}{25}=\frac{14}{25},\qquad \text{S.D.}=\frac{\sqrt{14}}{5}\approx0.75.\]

Answer (ii): Mean \(=\dfrac65\), Variance \(=\dfrac{14}{25}\), S.D. \(=\dfrac{\sqrt{14}}5\approx0.75\).

Q75 Marks

The probability distribution of a random variable \(X\) is as follows:

\(x\)	1	2	3	4	5	6
\(f(x)\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)

If \(y=2x+3\), find the expectation, variance and standard deviation of the random variable \(Y\).

Solution

Step 1: Mean and variance of X\[E(X)=\frac{1+2+3+4+5+6}{6}=\frac72,\qquad E(X^2)=\frac{1+4+9+16+25+36}{6}=\frac{91}6.\]\[\mathrm{Var}(X)=\frac{91}6-\left(\frac72\right)^2=\frac{91}6-\frac{49}4=\frac{182-147}{12}=\frac{35}{12}.\]Step 2: Mean and variance of Y = 2X + 3

Using \(E(aX+b)=aE(X)+b\) and \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\) with \(a=2,\ b=3\):

\[E(Y)=2\cdot\frac72+3=10.\]\[\mathrm{Var}(Y)=2^2\cdot\frac{35}{12}=\frac{35}3.\]\[\text{S.D. of }Y=\sqrt{\frac{35}3}\approx3.42.\]

Answer: \(E(Y)=10\), \(\mathrm{Var}(Y)=\dfrac{35}3\), S.D. \(=\sqrt{\dfrac{35}3}\approx3.42\).

Q85 Marks

Let \(x\) and \(y\) be two independent random variables having variances \(k\) and 2 respectively. If the variance of \(z=3y-x\) is 25, find \(k\).

Solution

For independent random variables, \(\mathrm{Var}(aX+bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)\). Here \(z=(-1)x+3y\), so \(a=-1\) and \(b=3\):

\[\mathrm{Var}(z)=(-1)^2\mathrm{Var}(x)+3^2\mathrm{Var}(y)=k+9\times2=k+18.\]

Given \(\mathrm{Var}(z)=25\):

\[k+18=25\ \Rightarrow\ k=7.\]

Answer: \(k=7\).

Q95 Marks

For a random variable \(X\), it is given \(E(X)=10\) and \(\mathrm{Var}(X)=25\). Find the positive values of \(a\) and \(b\) such that \(Y=aX-b\) has expectation 0 and variance 1.

Solution

Using \(E(aX+c)=aE(X)+c\) and \(\mathrm{Var}(aX+c)=a^2\mathrm{Var}(X)\) with \(c=-b\):

Variance condition\[\mathrm{Var}(Y)=a^2\,\mathrm{Var}(X)=25a^2=1\ \Rightarrow\ a^2=\frac1{25}\ \Rightarrow\ a=\frac15\quad(a>0).\]Expectation condition\[E(Y)=aE(X)-b=10a-b=0\ \Rightarrow\ b=10a=10\times\frac15=2.\]

Answer: \(a=\dfrac15\) and \(b=2\).

Q105 Marks

A number is chosen at random from the set \(1,2,3,\dots,150\) and another number is chosen at random from the set \(1,2,3,\dots,75\). Find the expected value of the product of the chosen numbers.

Solution

Let \(X\) be the number chosen from \(\{1,\dots,150\}\) and \(Y\) the number chosen from \(\{1,\dots,75\}\). The two choices are independent, so \(E(XY)=E(X)\,E(Y)\).

Expectation of X

Each of the 150 numbers has probability \(\frac1{150}\):

\[E(X)=\frac1{150}(1+2+\cdots+150)=\frac1{150}\cdot\frac{150\times151}{2}=\frac{151}{2}=75.5.\]Expectation of Y\[E(Y)=\frac1{75}(1+2+\cdots+75)=\frac1{75}\cdot\frac{75\times76}{2}=\frac{76}{2}=38.\]Expected value of the product\[E(XY)=E(X)E(Y)=\frac{151}2\times38=151\times19=2869.\]

Answer: The expected value of the product is 2869.

Formulas used: \(E(X)=\sum xp(x)\); \(\mathrm{Var}(X)=E(X^2)-\{E(X)\}^2\); \(E(aX+b)=aE(X)+b\); \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\); for independent \(X,Y\): \(\mathrm{Var}(aX+bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)\) and \(E(XY)=E(X)E(Y)\). A game is fair when each player's expectation is the same.