The problem asks for the probability that either A or B wins the game but not both. Let’s break it down step-by-step.

Notation:

• Probability of A winning the game, $P(A) = \frac{1}{3}$
• Probability of B winning the game, $P(B) = \frac{2}{5}$
• The probability of A not winning, $P(A^c) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$
• The probability of B not winning, $P(B^c) = 1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5}$

We want to find the probability that only one of them wins, i.e., either A wins and B loses, or B wins and A loses.

Case 1: A wins, B loses

The probability for this scenario is $P(A) \times P(B^c)$: $P(A \text{ wins and } B \text{ loses}) = P(A) \times P(B^c) = \frac{1}{3} \times \frac{3}{5} = \frac{1}{5}$

Case 2: B wins, A loses

The probability for this scenario is $P(B) \times P(A^c)$: $P(B \text{ wins and } A \text{ loses}) = P(B) \times P(A^c) = \frac{2}{5} \times \frac{2}{3} = \frac{4}{15}$

Total Probability

Since these two cases are mutually exclusive (either A wins and B loses or B wins and A loses), we can add the probabilities: $P(\text{Only one wins}) = \frac{1}{5} + \frac{4}{15}$ To add these, we need a common denominator: $\frac{1}{5} = \frac{3}{15}$ So, $P(\text{Only one wins}) = \frac{3}{15} + \frac{4}{15} = \frac{7}{15}$

Thus, the probability that only one of them wins is $\frac{7}{15}$.

Would you like further clarification or more details?

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Here are 5 related questions:

1. What is the probability that neither A nor B wins?
2. What is the probability that both A and B win?
3. If the probability of A winning increased to $\frac{1}{2}$, how would the result change?
4. How do probabilities change if A and B are dependent events?
5. What is the complementary probability that both either win or lose together?

Tip: Probabilities are always between 0 and 1. When you add probabilities, make sure the events are mutually exclusive!