Question

A bag contains two white balls and $3$ black balls. Four persons $A,B,C$ and $D$ in this order, draw a ball from the bag without replacement. The first person to draw a white ball is to receive Rs.$20.$ Their respective expectations are

• A. $E\left(A\right)=8,E\left(B\right)=6,E\left(C\right)=4,E\left(D\right)=2$
• B. $E\left(A\right)=6,E\left(B\right)=4,E\left(C\right)=8,E\left(D\right)=2$
• C. $E\left(A\right)=4,E\left(B\right)=6,E\left(C\right)=2,E\left(D\right)=8$
• D. $E\left(A\right)=8,E\left(B\right)=4,E\left(C\right)=6,E\left(D\right)=2$

Answer 1

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

Probability of drawing a white ball by $A=\frac{2}{2+3}=\frac{2}{5}$
$\therefore$ Amount expected by $A=\frac{2}{5}\times 20=8$
$B$ will draw a white ball only if $A$ draws a black ball.
$\therefore$ Probability of drawing a white ball by $B=\frac{3}{5}\times \frac{2}{4}=\frac{3}{10}$
$\therefore$ Amount expected by $B=\frac{3}{10}\times 20=6$
$C$ will draw a white ball only if both $A$ and $B$ draw black balls.
$\therefore$ Probability of drawing a white ball by $C=\frac{3}{5}\times \frac{2}{4}\times \frac{2}{3}=\frac{1}{5}$
$\therefore$ Amount expected by $C=\frac{1}{5}\times 20=4$
$D$ will draw a white ball only if $A,B$ and $C$ draw black ball each.
$\therefore$ Probaility 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 $D=\frac{1}{10}\times 20=2$