Three boys play a game of luck in which their respective chances of winning are \(\frac{1}{2}\), \(\frac{1}{3}\), and \(\frac{1}{4}\). What is the probability that one and only one of the boys wins the game?
The probabilities of the three boys winning are independent and are given as:
- Boy A: \( \frac{1}{2} \), - Boy B: \( \frac{1}{3} \), - Boy C: \( \frac{1}{4} \)
"One and only one wins" means exactly one boy wins (the other two lose).
The three possible cases are:
A wins, B loses, C loses
\( P = \frac{1}{2} \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} \)
B wins, A loses, C loses
\( P = \left(1 - \frac{1}{2}\right) \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} \)
C wins, A loses, B loses
\( P = \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \frac{1}{4} = \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} = \frac{2}{24} \)
Add them up:
\( \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24} \)
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