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: \( \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24} \)