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 expectation

• A. $E\left(A\right)=Rs.10,E\left(B\right)=Rs.6,E\left(C\right)=Rs.3,E\left(D\right)=Rs.1$
• B. $E\left(A\right)=Rs.8,E\left(B\right)=Rs.7,E\left(C\right)=Rs.4,E\left(D\right)=Rs.1$
• C. $E\left(A\right)=Rs.8,E\left(B\right)=Rs.6,E\left(C\right)=Rs.4,E\left(D\right)=Rs.2$
• D. $E\left(A\right)=Rs.10,E\left(B\right)=Rs.5,E\left(C\right)=Rs.3,E\left(D\right)=Rs.2$

Answer 1

The correct option is C $E\left(A\right)=Rs.8,E\left(B\right)=Rs.6,E\left(C\right)=Rs.4,E\left(D\right)=Rs.2$

Probability of drawing a white ball by $A=\frac{2}{2+3}=\frac{2}{5}$
Therefore amount expected by $A=\frac{2}{5}\times 20=Rs8$
B will draw a white ball only if $A$ draws a white ball.
Therefore probability of drawing a white ball by $B=\frac{3}{5}\times \frac{2}{4}=\frac{3}{10}$
Since black ball drawn by $A$ is not replaces in the bag
Therefore amount expected by $B=\frac{3}{10}\times 20=Rs6$
$C$ will draw a white ball only if both $A$ and $B$ draw black ball
Therefore probability of drawing a white ball by $C=\frac{3}{5}\times \frac{2}{4}\times \frac{2}{3}=\frac{1}{5}$
Therefore probability of drawing a white ball by $C=\frac{1}{5}\times 20=Rs4$
$D$ will draw a white ball only $A,B,C$ draw black ball each.
Therefore probability of drawing a white ball by $D=\frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times \frac{2}{2}=\frac{1}{10}$
Therefore amount expected by $D=\frac{1}{10}\times 20=Rs2$