Question

A had in his pocket a 1 rupee coin and four 10 paise coin, taking out 2 coins at random he promises to give them to B and C. What is the worth of C's expectation?
A. 28 paise
B. 38 paise
C. 48 paise
D. 58 paise

Answer 1

Hint: C can get two coins from A which either 1 one rupee and 1 ten paise coin or 2 ten paise coins. Find the expectation for both the cases. Find the probabilities of A taking out two different or the same coins and the expectation of C getting the coins for each case. Then find cumulative expectation using the formula for expectation.

Complete step-by-step solution:
The expectation or the average value of a number of a discrete random variable with $n$ events say $X={{a}_{1}},X={{a}_{2}},...,X={{a}_{n}}$ with respective probabilities $P{{\left( X=a \right)}_{1}},P\left( X={{a}_{2}} \right),...,P\left( X={{a}_{n}} \right)$ and respective value of outcomes ${{A}_{1}},{{A}_{2}},...,{{A}_{n}}$ is denoted as $E\left( X \right)$and is given by
\[E\left( X \right)=P\left( X={{a}_{1}} \right){{A}_{1}}+P\left( X={{a}_{2}} \right){{A}_{2}}+...+P\left( X={{a}_{n}} \right){{A}_{n}}\]
We see in the given question that the event is the person A randomly choosing two coins. There is 1 one rupee coin and 4 ten paise coins. So there are total 5 coins out of which A will randomly choose 2 coins. So the number of ways A can choose 2 out of 5 coins is the size of sample space. We calculate it using combination formula ,$ ^{5}{{C}_{2}}=\dfrac{5!}{3!2!}=\dfrac{5.4}{1.2}=10.$\[\]
We also see that the chosen two coins may be 1 one rupee coin and 1 ten paise coin or 2 ten paise coin. We are asked to find the expectation for C . \[\]
Case-1: \[\]
If A chooses 1 one rupee(100 paise ) coin and 1 ten paise coin from 4 ten paise coins, A can do it in $1{{\times }^{4}}{{C}_{1}}=1\times 4=4$ ways and the probability of such an event is $P\left( X=100,Y=10 \right)=\dfrac{4}{10}=\dfrac{2}{5}$.\[\]
We see in case-1 that C can get a coin of 1 one rupee(100 paise) or 1 ten paise with probability $P\left( X=10 \right)=P\left( X=10 \right)=\dfrac{1}{2}$ . So the expectation for C getting a coin in case-1 is $P\left( X=100 \right)\times 100+P\left( X=10 \right)\times 10=\dfrac{1}{2}\times 100+\dfrac{1}{2}\times 10=55$\[\]
 Case-2:\[\]
 If A chooses 2 ten paise coin , A can do it in $^{4}{{C}_{2}}=\dfrac{4!}{2!2!}=6$ ways and the probability of such an event is $P\left( X=10,Y=10 \right)=\dfrac{6}{10}=\dfrac{3}{5}$ \[\]
We see in case-2 that C can get 2 ten paise coins with probability $P\left( X=10 \right)=\dfrac{1}{2}$ . So the expectation for C getting a coin in case-2 is $P\left( X=10 \right)\times 10+P\left( X=10 \right)\times 10=2\times \dfrac{1}{2}\times 10=10$\[\]
So the total expectation combining both the case is
\[{{E}_{c}}=P\left( X=100,Y=10 \right)\times 55+P\left( X=10,Y=10 \right)\times 10=\dfrac{2}{5}\times 55+\dfrac{3}{5}\times 10=28\]
So the correct option is A.

Note: We can see here we have to calculate the expectation two times because A getting the coins and C getting the coins is not the same event. We could ignore the expectation B because B and C each getting coins from A is equally likely(equally probable) events.