- #1
Somefantastik
- 226
- 0
My book uses this certain notation, but doesn't seem to explain it. It's probably something that I should already know...
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'})
= \frac{P(G)P(B^{c})}{(P(G)P(B^{c}) + P(B)P(G^{c})}
I think the reason I don't follow is a notation thing. Can I treat the P{G, not Bill} as P(G\wedgeB^{c})? 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) ?
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'})
= \frac{P(G)P(B^{c})}{(P(G)P(B^{c}) + P(B)P(G^{c})}
I think the reason I don't follow is a notation thing. Can I treat the P{G, not Bill} as P(G\wedgeB^{c})? 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) ?
