Target practice with two shooters: P(A hits) = 5/7 and P(B hits) = 7/10, independently. What is the probability exactly one of them hits?

Introduction / Context:We need the probability of the exclusive OR (XOR) event: exactly one out of two independent shooters hits the target.

Given Data / Assumptions:P(A) = 5/7 ⇒ P(Ā) = 2/7; P(B) = 7/10 ⇒ P(B̄) = 3/10; independence assumed.

Concept / Approach:Compute P(A ∩ B̄) + P(Ā ∩ B). Multiply due to independence and sum the disjoint cases.

Step-by-Step Solution:

Verification / Alternative check:Complement method: P(exactly one) = P(A ∪ B) − 2P(A ∩ B) = [5/7 + 7/10 − (5/7)(7/10)] − 2*(5/7)(7/10) = 29/70.

Why Other Options Are Wrong:31/70 or 33/70 come from arithmetic slips; 21/10 is impossible (>1).

Common Pitfalls:Using “at least one” instead of “exactly one,” or failing to apply independence correctly.

Final Answer:29/70