Solving Notation Question: Probability of G's Shot

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SUMMARY

The discussion centers on calculating probabilities related to target shooting scenarios involving two shooters, Bill and George. Bill has a hit probability of 0.7, while George has a hit probability of 0.4. The key calculations involve determining the probability that George's shot was the only hit given that exactly one shot hit the target, as well as the probability that George's shot was the hit given that the target was hit. The correct formulas for these probabilities utilize conditional probability and the independence of events.

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  • Understanding of conditional probability
  • Familiarity with independent events in probability theory
  • Knowledge of basic probability notation (e.g., P(A), P(A ∩ B))
  • Ability to manipulate probability formulas
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  • Study the law of total probability
  • Learn about Bayes' theorem and its applications
  • Explore examples of independent events in probability
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This discussion is beneficial for students of probability theory, educators teaching statistics, and anyone interested in enhancing their understanding of conditional probabilities and independent events.

Somefantastik
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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'})

= [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) ?
 
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P{G, not Bill} = P(G and B') is correct.

I don't understand your last question.
 

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