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

# Barbara and Dianne go target shooting. Suppose that each of Barbara’s shots hits a wooden duck target with probability p1

J
240 points

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?
What independence assumptions have you made?

Barbara and
jeffp

8.7k points

Let,

H be the event that the duck is hit

B be the event that Barbra hit the duck

D be the event that Dianne hit the duck

both shots hit the duck

we want to find

$$P(BD|H)=\frac{P(H|BD)P(BD)}{P(H)}=\frac{P(BD)}{P(H)}$$

the probability P(H) is the probability that at least one person hits the target ie,

$$P(H)=P(BD^c)+P(B^cD)+P(BD)$$

$$=p_1(1-p_2)+(1-p_1)p_2+p_1p_2$$

therefore

$$P(BD|H)=\frac{P(BD)}{P(H)}$$

$$=\frac{p_1p_2}{p_1(1-p_2)+(1-p_1)p_2+p_1p_2}$$

$$=\frac{p_1p_2}{p_1+p_2-p_1p_2}$$

Barbara's shot hit the duck

This happens under two scenarios, Barbara alone hits the duck, or both Barbara and Diane hit the duck. The second scenario was already calculated in part (a). So we have that,

$$P(B|H)=P(BD^c|H)+P(BD|H)$$

we already know $P(BD|H)$, but need to find

$$P(BD^c|H)=\frac{p_1(1-p_2)}{p_1+p_2-p_1p_2}$$

and therefore

$$P(B|H)=P(BD^c|H)+P(BD|H)$$

$$=\frac{p_1p_2}{p_1+p_2-p_1p_2}+\frac{p_1(1-p_2)}{p_1+p_2-p_1p_2}$$

$$=\frac{p_1}{p_1+p_2-p_1p_2}$$

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