91Ó°ÊÓ

Since there are 4 white balls and 12 balls in total, the probability of player A drawing a white ball in a single turn is \(\frac{4}{12} = \frac{1}{3}\).

Player B can only win if player A loses in their turn. The probability of player A losing in one turn is \(1 - \frac{1}{3} = \frac{2}{3}\). If A loses, then the probability of player B drawing a white ball is still \(\frac{1}{3}\). Thus, the probability of player B winning in one turn is \(\frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}\).

Player C can only win if both player A and B lose in their turns. The probability of player A and B both losing in one turn is \(\frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}\). If both A and B lose, the probability of player C drawing a white ball is \(\frac{1}{3}\). Thus, the probability of player C winning in one turn is \(\frac{4}{9} \cdot \frac{1}{3} = \frac{4}{27}\).

Now that we have the probabilities for each player winning in one turn, we can find the probability of each player winning by considering the geometric series. Let the probabilities for player A, B, and C winning be \(P_A, P_B\), and \(P_C\), respectively. For player A: \(P_A = \frac{1}{3} + \frac{2}{9}\cdot\frac{4}{27} + \left(\frac{2}{9}\cdot\frac{4}{27}\right)^2 + \ldots\) For player B: \(P_B = \frac{2}{9} + \frac{2}{9}\cdot\frac{4}{27} + \left(\frac{2}{9}\cdot\frac{4}{27}\right)^2 + \ldots\) For player C: \(P_C = \frac{4}{27} + \left(\frac{2}{9}\cdot\frac{4}{27}\right)^2 + \ldots\) The geometric series can be used to calculate the sum, and the sum can be represented as \(\frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio in each series. For player A: \(P_A = \frac{\frac{1}{3}}{1-\frac{8}{27}} = \frac{9}{19}\) For player B: \(P_B = \frac{\frac{2}{9}}{1-\frac{8}{27}} = \frac{6}{19}\) For player C: \(P_C = \frac{\frac{4}{27}}{1-\frac{8}{27}} = \frac{4}{19}\) So, the probabilities of winning for players A, B, and C are \(\frac{9}{19}, \frac{6}{19}\), and \(\frac{4}{19}\) respectively when the balls are replaced after being drawn. (b) Balls are not replaced after being drawn

With the balls not replaced, the probability of player A drawing a white ball in one turn is the same as before, \(\frac{4}{12} = \frac{1}{3}\).

Player B can only win if player A loses in their turn. The probability of player A losing in one turn is \(1 - \frac{1}{3} = \frac{2}{3}\). If A loses, there are now 11 balls left, and the probability of player B drawing a white ball is \(\frac{4}{11}\). Thus, the probability of player B winning in one turn is \(\frac{2}{3} \cdot \frac{4}{11} = \frac{8}{33}\).

Player C can only win if both player A and B lose in their turns. The probability of player A and B both losing in one turn is \(\frac{2}{3} \cdot \frac{7}{11} = \frac{14}{33}\). If both A and B lose, there are now 10 balls left, and the probability of player C drawing a white ball is \(\frac{4}{10} = \frac{2}{5}\). Thus, the probability of player C winning in one turn is \(\frac{14}{33} \cdot \frac{2}{5} = \frac{28}{165}\).

Let the probabilities for player A, B, and C winning be \(P_A, P_B\), and \(P_C\), respectively. For player A: \(P_A = \frac{1}{3}\) For player B: \(P_B = \frac{8}{33}\) For player C: \(P_C = 1 - P_A - P_B = 1 - \frac{1}{3} - \frac{8}{33} = \frac{28}{99}\) So, the probabilities of winning for players A, B, and C are \(\frac{1}{3}, \frac{8}{33}\), and \(\frac{28}{99}\) respectively when the balls are not replaced after being drawn.