Two balls are chosen randomly from an urn containing 8 white and 4 black balls by a player. Suppose that he wins Rs. 30 for each black ball selected and loses Rs. 15 for each white ball selected. The expected value of winning amount is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Rs. 0
Let X be the amount won (in Rs.). The number of ways of choosing 2 balls out of 12 is \({}^{12}C_{2} = 66\).
2 white: \(X = -15 - 15 = -30\), ways \({}^{8}C_{2} = 28\)
1 white, 1 black: \(X = -15 + 30 = 15\), ways \(8 \times 4 = 32\)
2 black: \(X = 30 + 30 = 60\), ways \({}^{4}C_{2} = 6\)
The expected value is \(E(X) = \sum x\,P(x)\):
\( \displaystyle E(X) = \frac{(-30)(28) + 15(32) + 60(6)}{66} \)
\( \displaystyle E(X) = \frac{-840 + 480 + 360}{66} = \frac{0}{66} = 0 \)
So the expected winning amount is Rs. 0: on average the player neither gains nor loses.