- #1

Somefantastik

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Bill and George (Noice!) go target shooting at each other. Both shoot at the target at the same time. Bill hits target with prob 0.7 and G independently hits target with prob 0.4.

(a) Given that exactly one shot hit the target, what is the prob it was G's shot?

[sol]

P{G|exactly 1 hit} = P{G, not Bill}/ P(exactly 1 hit}

= P{G, B'}/(P{G,B'} + P{B,G'})

= [tex]\frac{P(G)P(B^{c})}{(P(G)P(B^{c}) + P(B)P(G^{c})}[/tex]

I think the reason I don't follow is a notation thing. Can I treat the P{G, not Bill} as P(G[tex]\wedge[/tex]B[tex]^{c}[/tex])? If that's the case then it makes perfect sense as the events are independent.

(b) Given that the target was hit, what's the probability it was G's shot?

P(G|H) = P(G)/P(H)

where

P(H) = 1- P(no hit) = 1 - (P(G')P(B'))

Why is it not P(G|H) = P(G)/P(G)P(B) ?