Question

Hunters A and B shoot at a target; the probabilities of hitting the target are \(p_{1}\) and \(p_{2}\), respectively. Assuming independence, can \(p_{1}\) and \(p_{2}\) be selected so that \(P(\) zero hits \()=P(\) one hit \()=P(\) two hits \() ?\)

Short answer

Yes, both \(p_{1}\) and \(p_{2}\) can be \(\frac{1}{3}\), which makes the probabilities for zero hits, one hit, and two hits all equal.

Step-by-step solution

Zero hits

The probability of zero hits is the probability that both hunters miss the target. This is expressed as \((1 - p_{1})(1 - p_{2})\).

One hit

The probability of one hit is the sum of the probabilities that one hunter hits the target and the other misses, i.e. the hunter A hits and the hunter B misses, or the hunter A misses and the hunter B hits. This is expressed as \(p_{1}(1-p_{2}) + (1 - p_{1})p_{2}\).

Two hits

The probability of two hits is the probability that both hunters hit the target. This is expressed as \(p_{1}p_{2}\).

Solve the equations

We can now set the equations equal to each other: \((1 - p_{1})(1 - p_{2}) = p_{1}(1-p_{2}) + (1 - p_{1})p_{2} = p_{1}p_{2}\). After some computation, we find that \(p_{1}\) and \(p_{2}\) can each be \(\frac{1}{3}\). This satisfies all three conditions for zero hits, one hit, and two hits.