Question

(a) A bag contains two white balls and 3 black balls. Four person 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. (a) E (A) =Rs 8, E (B) = Rs.6, E(C) = Rs.4 and E (D) = Rs 2
• B. (a) E (A) =Rs 6, E (B) = Rs.4, E(C) = Rs.8 and E (D) = Rs 2
• C. (a) E (A) =Rs 4, E (B) = Rs.6, E(C) = Rs.2 and E (D) = Rs 8
• D. (a) E (A) =Rs 8, E (B) = Rs.4, E(C) = Rs.6 and E (D) = Rs 2

Answer 1

Answer: (A)

(a) E (A) =Rs 8, E (B) = Rs.6, E(C) = Rs.4 and E (D) = Rs 2

$\therefore$ 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 red ball.
$\therefore$ Probability of drawing a white ball by $B=\frac{3}{5}\times \frac{2}{4}=\frac{3}{10}$
[$\because$ black ball drawn by $A$ is not replaced 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 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 $C=\frac{1}{5}\times 20=Rs4$
$D$ will draw a white ball only if $A,B,C$ 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 $D=\frac{1}{10}\times 20=Rs2$