A bag contains 2 red, 3 green and 5 white balls. A player pays Rs. k to draw one ball at random. He gets Rs. 50 if the ball is red, Rs. 20 if it is green and nothing if it is white. The game is fair (the expected net gain of the player is zero) if k equals
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 16
Let X be the player’s net gain (in Rs.), that is, the prize minus the fee k:
\(P(\text{red}) = 0.2,\quad P(\text{green}) = 0.3,\quad P(\text{white}) = 0.5\)
\(E(X) = (50 - k)(0.2) + (20 - k)(0.3) + (0 - k)(0.5)\)
\(= 10 + 6 - k\)
\(= 16 - k\)
A fair game has expected net gain 0:
\(16 - k = 0\), so \(k = 16\)
So the fair fee equals the expected prize, Rs. 16.
Why the other options are incorrect
Option (1) 10 — Incorrect. This counts only the red prize, \(50 \times 0.2\).
Option (2) 35 — Incorrect. This is the plain average of the two prizes, \(\frac{50 + 20}{2}\), ignoring their probabilities.
Option (3) 6 — Incorrect. This counts only the green prize, \(20 \times 0.3\).