Question

A had in his pocket a rupee and four 10 paise coins, taking out two coins at random he promises to give them to B and C. What is the worth of B’s expectation?
(a) 28 Paise
(b) 38 Paise
(c) 48 Paise
(d) 58 Paise

Answer 1

There are a total of 5 coins. Firstly, we have to find the number of ways to choose 2 coins from 5 coins which will be $^{5}{{C}_{2}}$ . A can give 2 coins to B in two cases. A can give either 1 rupee coin (100 paise) or one 10 paise coin from four 10 paise coins in $1{{\times }^{4}}{{C}_{1}}$ . We have to find the probability of this and the expectation. Then, we have to find the expectation and probability for the case A can give two 10 paise coins from four 10 paise coins. Finally, find the cumulative expectation using the formula.

Complete step by step answer:
We know that the expectation (average value) of a number of discrete random variable with n events $X={{a}_{1}},X={{a}_{2}},...,X={{a}_{n}}$ with probabilities $P\left( X={{a}_{1}} \right),P\left( X={{a}_{2}} \right),...,P\left( X={{a}_{n}} \right)$ and outcomes ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ is given by
$E\left( X \right)=P\left( X={{a}_{1}} \right){{x}_{1}}+P\left( X={{a}_{2}} \right){{x}_{2}}+...+P\left( X={{a}_{n}} \right){{x}_{n}}$
We are given that A had in his pocket a rupee and four 10 paise coins. Therefore, there are total of 5 coins. We are given that A has taken out 2 coins randomly. We know that the number of ways to choose r items from n items is given by $^{n}{{C}_{r}}$ . Therefore, the number of ways to choose 2 coins from 5 coins is
${{\Rightarrow }^{5}}{{C}_{2}}$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ .
$\begin{aligned} & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{5!}{\left( 5-2 \right)!2!} \\\ & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{5!}{3!2!} \\\ \end{aligned}$
We know that $n!=n\left( n-1 \right)!$ .
$\begin{aligned} & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{5\times 4\times 3!}{3!2\times 1} \\\ & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{5\times 4\times \require{cancel}\cancel{3!}}{\require{cancel}\cancel{3!}2\times 1} \\\ & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{20}{2} \\\ & {{\Rightarrow }^{5}}{{C}_{2}}=10 \\\ \end{aligned}$
A can give 2 coins to B in following ways:
Case1: A can give either 1 rupee coin (100 paise) or one 10 paise coin from four 10 paise coins. He can do this in $1{{\times }^{4}}{{C}_{1}}$ ways.
Let us find the value of $1{{\times }^{4}}{{C}_{1}}$ .
$\Rightarrow 1{{\times }^{4}}{{C}_{1}}{{=}^{4}}{{C}_{1}}$
We know that $^{n}{{C}_{1}}=n$ .
$\Rightarrow 1{{\times }^{4}}{{C}_{1}}=4$
Now, let us find the probability of this event.
$\Rightarrow {{P}_{1}}\left( X=100,Y=10 \right)=\dfrac{4}{10}=\dfrac{2}{5}$
We can see that B can get a coin of 1 rupee (100 paise) or one 10 paise with probabilities given by
$\Rightarrow P\left( X=100 \right)=P\left( X=10 \right)=\dfrac{1}{2}$
Therefore, the expectation of B getting two coins as in case 1 is
$\begin{aligned} & \Rightarrow {{E}_{1}}\left( X \right)=P\left( X=100 \right)\times 10+P\left( X=10 \right)\times 10 \\\ & \Rightarrow {{E}_{1}}\left( X \right)=\dfrac{1}{2}\times 100+\dfrac{1}{2}\times 10 \\\ & \Rightarrow {{E}_{1}}\left( X \right)=55 \\\ \end{aligned}$
Case 2: A can give two 10 paise coins from four 10 paise coins. He can do this in $^{4}{{C}_{2}}$ ways.
$\begin{aligned} & {{\Rightarrow }^{4}}{{C}_{2}}=\dfrac{4!}{\left( 4-2 \right)!2!} \\\ & {{\Rightarrow }^{4}}{{C}_{2}}=\dfrac{4!}{2!2!} \\\ & {{\Rightarrow }^{4}}{{C}_{2}}=\dfrac{4\times 3\times 2\times 1}{4} \\\ & {{\Rightarrow }^{4}}{{C}_{2}}=6 \\\ \end{aligned}$

Now, let us find the probability of this event.
$\Rightarrow {{P}_{2}}\left( X=10,Y=10 \right)=\dfrac{6}{10}=\dfrac{3}{5}$
We can see that B can get two 10 paise with probabilities given by

& \Rightarrow P\left( X=10\text{ paise} \right)=P\left( X=10\text{ paise} \right)=\dfrac{1}{2} \\\ & \Rightarrow P\left( X=10 \right)=P\left( X=10 \right)=\dfrac{1}{2} \\\ \end{aligned}$$ Therefore, the expectation of B getting two coins as in case 2 is $\begin{aligned} & \Rightarrow {{E}_{2}}\left( X \right)=P\left( X=10 \right)\times 10+P\left( X=10 \right)\times 10 \\\ & \Rightarrow {{E}_{2}}\left( X \right)=\dfrac{1}{2}\times 10+\dfrac{1}{2}\times 10 \\\ & \Rightarrow {{E}_{2}}\left( X \right)=10 \\\ \end{aligned}$ Therefore, the total expectation can be found as follows. $$\begin{aligned} & \Rightarrow E\left( X \right)={{P}_{1}}\left( X=100,Y=10 \right)\times {{E}_{1}}\left( X \right)+{{P}_{2}}\left( X=10,Y=10 \right)\times {{E}_{2}}\left( X \right) \\\ & \Rightarrow E\left( X \right)=\dfrac{2}{5}\times 55+\dfrac{3}{5}\times 10 \\\ & \Rightarrow E\left( X \right)=22+6 \\\ & \Rightarrow E\left( X \right)=28\text{ paise} \\\ \end{aligned}$$ **So, the correct answer is “Option a”.** **Note:** Students must be thorough with the formulas of probability, expectation, variance and standard deviation. They must never forget to convert the given rupees into paise since the final answer must be in paise. The expectation of B getting the 2 coins will be the same as C getting since both these events are equally likely.