Question

A bag contains 2 white balls and 3 black balls. Four persons A, B, C, 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 the expectations.
A. $E\left( A \right)=10;E\left( B \right)=6;E\left( C \right)=3;E\left( D \right)=1$
B. $E\left( A \right)=8;E\left( B \right)=7;E\left( C \right)=4;E\left( D \right)=1$
C. $E\left( A \right)=8;E\left( B \right)=6;E\left( C \right)=4;E\left( D \right)=2$
D. $E\left( A \right)=10;E\left( B \right)=5;E\left( C \right)=3;E\left( D \right)=2$

Answer 1

Hint: We first form the events for all four people A, B, C, D who were drawing a ball from the bag and not replace it. We find the probabilities for all the events and for the individuals also. We multiply the probability with the winnable amount to find the expectations. So, for A, we will define the event as A and we find the probability as $\dfrac{2}{2+3}=\dfrac{2}{5}$. Then we will multiply this and Rs. 20 to get the expected amount for A. Similarly, we can proceed for B, C and D.

Complete step-by-step answer:
We need the expectations for four persons A, B, C, D who were drawing a ball from the bag and not replacing it.
In the bag there are 2 white balls and 3 black balls.
We know the exception for a variable x can be estimated as the formula
$E\left( x \right)=xP\left( x \right)$, here $P\left( x \right)$ defines the probability of x happening.
The order for the four people drawing the balls is also A, B, C, D.
So, they can withdraw balls for maximum 5 times.
Now, A starts and he has 2 white balls and 3 black balls at the start.
The probability of drawing a white ball by A is $\dfrac{2}{2+3}=\dfrac{2}{5}$.
The total amount winnable is Rs. 20.
So, the expected amount for A is $20\times \dfrac{2}{5}=8$.
The game will go to B only if A picks black ball at the start.
So, the events are A picks black ball and B picks white ball.
The probability of that event happening is $\dfrac{3}{5}\times \dfrac{2}{4}=\dfrac{6}{20}$.
So, the expected amount for B is $20\times \dfrac{6}{20}=6$.
The game will go to C only if both A and B pick black ball at the start.
So, the events are A and B both pick black balls and C picks white balls.
The probability of that event happening is $\dfrac{3}{5}\times \dfrac{2}{4}\times \dfrac{2}{3}=\dfrac{1}{5}$.
So, the expected amount for C is $20\times \dfrac{1}{5}=4$.
The game will go to D only if A, B, C all pick black ball at the start.
So, the events are A, B, C pick black balls and D picks white balls.
The probability of that event happening is $\dfrac{3}{5}\times \dfrac{2}{4}\times \dfrac{1}{3}\times 1=\dfrac{1}{10}$.
So, the expected amount for B is $20\times \dfrac{1}{10}=2$.
The correct option is C.

So, the correct answer is “Option C”.

Note: When we were finding the probabilities, we were taking the ratio of picking a specific coloured ball and the total number of balls remaining. All the events of picking a ball for an individual are independent to the other one and that’s why we were multiplying them to find the expectations for every single person.