Question

Two balls are chosen randomly from an urn containing 8 white balls and 4 black balls by a player. Suppose that he wins ₹30 for each black ball selected and loses ₹15 for each white ball selected. The expected value of winning amount is :

• A. ₹12.72
• B. ₹14.72
• C. ₹15
• D. ₹00

Answer 1

The Correct Option is D

To solve the problem, we need to find the expected value of the amount a player wins by randomly choosing two balls from the urn containing 8 white balls and 4 black balls.

The possible scenarios when two balls are selected are:

1. Selecting 2 black balls.

2. Selecting 1 black ball and 1 white ball.

3. Selecting 2 white balls.

Let's compute the probability and winnings for each scenario:

1. Two black balls:

Total ways to choose any 2 balls: $C(12,2)=66$

Ways to choose 2 black balls: $C(4,2)=6$

Probability of 2 black balls: $\frac{6}{66}=\frac{1}{11}$

Winnings: $60$ (₹30 per black ball x 2)

2. One black ball and one white ball:

Ways to choose 1 black and 1 white ball: $C(4,1)C(8,1)=32$

Probability of 1 black and 1 white ball: $\frac{32}{66}=\frac{16}{33}$

Winnings: $15$ (₹30 for black - ₹15 for white)

3. Two white balls:

Ways to choose 2 white balls: $C(8,2)=28$

Probability of 2 white balls: $\frac{28}{66}=\frac{14}{33}$

Winnings: $-30$ (₹15 loss per white ball x 2)

Now, we calculate the expected value \(E\):

$\text{E}=\frac{1}{11}60+\frac{16}{33}15+\frac{14}{33}-30$

$=\frac{60}{11}+\frac{240}{33}-\frac{420}{33}$

Converting the fractions to a common denominator, we have:

$=\frac{180}{33}+\frac{240}{33}-\frac{420}{33}$

$=\frac{\mathrm{180+240-420}}{33}$

$=\frac{0}{33}$

The expected value is ₹0.