Question

Barbara and Dianne go target shooting. Suppose that each of Barbara’s shots hits a wooden duck target with probability p₁, while each shot of Dianne’s hits it with probability p₂. Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that

(a) both shots hit the duck?

(b) Barbara’s shot hit the duck?

Short answer

1. The probability that both shots hit the duck will be $\frac{p_1{p}_2}{1-q_1{q}_2}$.
2. The probability of Barbara’s shot hit the duck is$=\frac{p_1}{1-q_1{q}_2}$.

Step-by-step solution

Given information (Part a)

Suppose that each of Barbara’s shots hits a wooden duck target with probability p₁, while each shot of Dianne’s hits it with probability p₂.

Solution (Part a)

If the wooden duck is knocked over then the probability that both shots hit the duck is given by

$P\left[\text{Both shots hits the duck |at least one hit (the wooden duck is knocked over)}\right]$

$=\frac{P\left[\text{Both hits}\right]}{P\left[\text{at least one hits}\right]}$

$=\frac{P\left[\text{Both hits}\right]}{1-P\left[\text{no hits}\right]}$

$=\frac{p_1{p}_2}{1-q_1{q}_2}$

Final answer (Part a)

The probability that both shots hit the duck will be$\frac{p_1{p}_2}{1-q_1{q}_2}$.

Given information (Part b)

Barbara and Dianne go target shooting. Suppose that each of Barbara’s shots hits a wooden duck target with probability p₁, while each shot of Dianne’s hits it with probability p₂.

Solution (Part b)

If the wooden duck has knocked over then the probability that Barbara's shot struck the duck is shown below,

$P\left[\text{Barbara shot hit the duck |at least one hit(the wooden duck is knocked over)}\right]$

$=\frac{P\left[\text{Barbar hits}\right]}{P\left[\text{at least one hits}\right]}$

$=\frac{P\left[\text{Barbara hits}\right]}{1-P\left[\text{no hits}\right]}$

$=\frac{p_1}{1-q_1{q}_2}$

Final answer (Part b)

The probability of Barbara’s shot hit the duck is$=\frac{p_1}{1-q_1{q}_2}$.