Three playerslong dashA, B, and Clong dasheach have their own urn that contains 28 balls, of which 20 are white. The three players successively draw from their urns, A first, then B, then C, then A, and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player if
(a) each ball is replaced after it is drawn;
(b) the balls that are withdrawn are not replaced.

Question

10-05-2024
Mathematics

Answered

Three playerslong dashA, B, and Clong dasheach have their own urn that contains 28 balls, of which 20 are white. The three players successively draw from their urns, A first, then B, then C, then A, and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player if
(a) each ball is replaced after it is drawn;
(b) the balls that are withdrawn are not replaced.

Answer :

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e3van22 e3van22 · Answer 1 · 2024-05-10T22:03:32-04:00

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(b) The balls that are withdrawn are not replaced: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.