In a shooting competition, the probabilities of hitting the target by p, q and…

Concept: When several events are independent, the probability that exactly one of them occurs equals the sum, over each event, of that event's own probability of happening multiplied by the probabilities of every other event NOT happening. The three ways in which exactly one event can occur are mutually exclusive, so their probabilities simply add.

Application:

1. Let P(p hits) = 3/5, so P(p misses) = 1 - 3/5 = 2/5.

2. Let P(q hits) = 1/3, so P(q misses) = 1 - 1/3 = 2/3.

3. Let P(r hits) = 2/5, so P(r misses) = 1 - 2/5 = 3/5.

4. Case A - only p hits: P(p hits) x P(q misses) x P(r misses) = (3/5) x (2/3) x (3/5) = 18/75.

5. Case B - only q hits: P(p misses) x P(q hits) x P(r misses) = (2/5) x (1/3) x (3/5) = 6/75.

6. Case C - only r hits: P(p misses) x P(q misses) x P(r hits) = (2/5) x (2/3) x (2/5) = 8/75.

7. Since Cases A, B and C cannot happen together, add them: 18/75 + 6/75 + 8/75 = 32/75.

Cross-check: The probabilities of getting 0, 1, 2 and 3 hits must add up to 1.

• P(0 hits) = P(p misses) x P(q misses) x P(r misses) = (2/5) x (2/3) x (3/5) = 12/75.

• P(2 hits) = (3/5)x(1/3)x(3/5) + (3/5)x(2/3)x(2/5) + (2/5)x(1/3)x(2/5) = 9/75 + 12/75 + 4/75 = 25/75.

• P(3 hits) = P(p hits) x P(q hits) x P(r hits) = (3/5) x (1/3) x (2/5) = 6/75.

Adding all four: P(0 hits) + P(1 hit) + P(2 hits) + P(3 hits) = 12/75 + 32/75 + 25/75 + 6/75 = 75/75 = 1, confirming the value obtained for exactly one hit is consistent with the full probability distribution.

Result: Adding the three mutually exclusive cases from the Application step gives the probability that exactly one of them hits the target.