A hits a target 3 times in 5 shots and B hits it 2 times in 4 shots. If both fire once, what is the probability that the target is hit?
SHORT ANSWER
4/5. P(A hits) = 3/5, P(B hits) = 1/2. The target is missed only if both miss: 2/5 x 1/2 = 1/5. So P(hit) = 1 - 1/5 = 4/5.
Explain it simply
"Hit at least once" covers many cases: only A hits, only B hits, or both hit. The opposite has just one case: both miss. So find the chance of the single opposite case and subtract from 1. This complement trick is the fastest route for most "at least one" questions.
Example
Question: P(A hits) = 3/5, P(B hits) = 2/4 = 1/2. Both fire once. P(target hit)?
- P(A misses) = 2/5. P(B misses) = 1/2.
- Shots are independent, so P(both miss) = 2/5 x 1/2 = 1/5.
- P(hit) = 1 - 1/5 = 4/5.
Follow-ups:
- P(exactly one hits) = (3/5)(1/2) + (2/5)(1/2) = 3/10 + 2/10 = 1/2.
- P(both hit) = 3/5 x 1/2 = 3/10.
- Check: 1/2 + 3/10 = 8/10 = 4/5.
js
const a = 3 / 5, b = 1 / 2;
console.log(1 - (1 - a) * (1 - b)); // Output: 0.8What to say in the interview
"The only way the target survives is if both miss, which is 2/5 times 1/2, or 1/5. So the probability it is hit is 4/5. Exactly one hit plus both hit also adds to 4/5."
Common mistakes
- Adding 3/5 + 1/2 = 11/10, which is more than 1.
- Assuming the events are mutually exclusive; both can hit.