Question

A bag contains two white balls and 3 black balls. Four people A, B, C and D in this order, draw a ball from the bag and do not replace it. The first person to draw a white ball is to receive Rs. 20. Determine their expectations.
(a) E(A) = Rs. 8, E(B) = Rs. 6, E(C) = Rs. 4 and E(D) = Rs. 2
(b) E(A) = Rs. 6, E(B) = Rs. 4, E(C) = Rs. 8 and E(D) = Rs. 2
(c) E(A) = Rs. 4, E(B) = Rs. 6, E(C) = Rs. 2 and E(D) = Rs. 8
(d) E(A) = Rs. 8, E(B) = Rs. 4, E(C) = Rs. 6 and E(D) = Rs. 2

Answer 1

Hint: To solve the given question, we will first find out what probability is and what an exception is. Then, we will find the probabilities of A, B, C, and D that they are the first person to draw the white ball. If we consider probability of A, we have favourable outcomes as 2 and total outcomes as 2+3=5, so we will get \[P\left( A \right)=\dfrac{2}{5}\] . Similarly, we will find others too. Then, we will multiply their probabilities by the amount given to get their individual expectations.

Complete step by step solution:
Before we solve the given question, we must know what probability is and what an exception is. Probability is defined as the chances or possibilities of any random event. The expectation is defined as the product of the probability of any event and the value corresponding with the actual observed occurrence of the event. Thus, to find the expectations of A, B, C and D, we will first find their probabilities and then multiply them with the amount given.
Now, A draws a ball in the starting. There are five balls in total and two white balls. We know that the probability of any event X is given by
\[P\left( X \right)=\dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}}\]
In our case, favourable outcomes = 2 and total outcomes = 5. Therefore, we get,
\[P\left( A \right)=\dfrac{2}{5}\]
Now, the next ball is drawn by B. Now, if B is the first person to draw a white ball then A should choose a black ball. The probability of this will be
\[\Rightarrow 1-P\left( A \right)\]
\[\Rightarrow 1-\dfrac{2}{5}\]
\[\Rightarrow \dfrac{3}{5}\]
Now, there are four balls left which are total outcomes and two balls are white which are favorable outcomes. Thus, the probability becomes \[\dfrac{2}{4}.\] Now, the total probability will be
\[P\left( B \right)=\dfrac{3}{5}\times \dfrac{2}{4}\]
\[\Rightarrow P\left( B \right)=\dfrac{3}{10}\]
Now, the next ball is drawn by C. Now, if C is the first person to draw a white ball then A and B should get black balls and C should choose one out of two balls in the remaining three balls. Thus, the total probability will be,
\[P\left( C \right)=\dfrac{3}{5}\times \dfrac{2}{4}\times \dfrac{2}{3}\]
\[\Rightarrow P\left( C \right)=\dfrac{1}{5}\]
Similarly, the total probability for D to pick up the first white ball will be,
\[P\left( D \right)=\dfrac{3}{5}\times \dfrac{2}{4}\times \dfrac{1}{3}\times \dfrac{2}{2}\]
\[\Rightarrow P\left( D \right)=\dfrac{1}{10}\]
Now, their expectations will be obtained by multiplying their probabilities with the total amount given. The amount given in the question is Rs. 20. Thus, we have,
\[E\left( A \right)=P\left( A \right)\times Rs.20\]
\[\Rightarrow E\left( A \right)=\dfrac{2}{5}\times Rs.20\]
\[\Rightarrow E\left( A \right)=Rs.8\]
Similarly, for B, we have
\[E\left( B \right)=P\left( B \right)\times Rs.20\]
\[\Rightarrow E\left( B \right)=\dfrac{3}{10}\times Rs.20\]
\[\Rightarrow E\left( B \right)=Rs.6\]
Similarly, we can get it for C as
\[E\left( C \right)=P\left( C \right)\times Rs.20\]
\[\Rightarrow E\left( C \right)=\dfrac{1}{5}\times Rs.20\]
\[\Rightarrow E\left( C \right)=Rs.4\]
Similarly, for D, we can write it as
\[E\left( D \right)=P\left( D \right)\times Rs.20\]
\[\Rightarrow E\left( D \right)=\dfrac{1}{10}\times Rs.20\]
\[\Rightarrow E\left( D \right)=Rs.2\]
Hence, option (a) is the right answer.

Note: In this question, the order in which the persons are going to draw a ball is an important factor in determining the expectation. If the order of A, B, C and D changes then their value of expectation changes. Another thing to note here is that no matter what the order is, the sum of expectations will always be equal to the total amount.