Let's calculate the probability of winning for each player under the given conditions:
(a) Each ball is replaced after it is drawn:
Under this condition, the probability of drawing a white ball in each turn for each player remains the same at \( \frac{20}{28} = \frac{5}{7} \).
Player A:
The probability of winning for A on the first draw is \( \frac{20}{28} = \frac{5}{7} \).
Player B:
The probability of winning for B on the second draw is also \( \frac{5}{7} \).
Player C:
The probability of winning for C on the third draw is also \( \frac{5}{7} \).
So, under condition (a), the probability of winning for each player is the same, which is \( \frac{5}{7} \).
(b) The balls that are withdrawn are not replaced:
Under this condition, the probability of drawing a white ball decreases with each successive draw.
Player A:
The probability of A winning on the first draw is \( \frac{20}{28} = \frac{5}{7} \).
Player B:
After A's draw, there are \(28 - 1 = 27\) balls left, of which \(20 - 1 = 19\) are white. So, the probability of B winning on the second draw is \( \frac{19}{27} \).
Player C:
After B's draw, there are \(27 - 1 = 26\) balls left, of which \(20 - 1 = 19\) are white. So, the probability of C winning on the third draw is \( \frac{19}{26} \).
The probability of A winning on the fourth draw would be \( \frac{19}{26} \), and so on.
