My Math Forum Bayes' Theorem - problem

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 May 2nd, 2018, 03:59 AM #1 Newbie   Joined: May 2018 From: Serbia Posts: 2 Thanks: 0 Bayes' Theorem - problem Two archers ($S_1$ and $S_2$) are shooting target. In period of time $t$ archer $S_1$ shoots $9$ arrows, while archer $S_2$ shoots $10$ arrows. It's known that from $10$ arrows archer $S_1$ hits target with $8$ arrows, and from $10$ arrows archer $S_2$ hits target with $7$ arrows. While shooting target was hit. What is probability that target was hit by archer $S_2$? I tried to solve this problem with Bayes' theorem and I did get right answer, which is $P(H_2|A)=0.49$ but in formula $P(H_2|A)=\frac{P(H_2)P(A|H_2)}{P(A)}$ where $P(A)$ is total probability of even $A$ (target was hit) occurring and $P(A)=1.42>1$ (which is not possible) Other parameters are: $P(H_1) = 0.9, P(H_2) = 1, P(A|H_1) = 0.8, P(A|H_2)=0.7$ Last edited by andrijaada99; May 2nd, 2018 at 04:02 AM.
 May 2nd, 2018, 11:01 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 please define the terms you are using. What do $H_1,~H_2,~A$ stand for? These additional parameters you list at the bottom. You are given these as part of the problem statement?
 May 2nd, 2018, 02:03 PM #3 Newbie   Joined: May 2018 From: Serbia Posts: 2 Thanks: 0 $A$ - event that archers will hit target Hypothesis: $H_1$ - event that archer $S_1$ will shoot target $H_2$ - event that archer $S_2$ will shoot target "Other parameters" are not given as part of the problem, they are ones that I calculated from the text of problem. I manange to find error. It was when I defined $P(H_1)$ and $P(H_2)$. Since $H_1 + H_2 = E \implies P(H_1) + P(H_2) = 1$ Now since there is $10$ arrows and $S_1$ shoots $9$ for some time $t$, and there are $2$ archers then, $$P(H_1) = \frac{\frac{9}{10}}{2} + x = 0.45 + x$$ and same for other archer: $$P(H_2) = \frac{\frac{10}{10}}{2} + x = 0.5 + x$$ Now we have: $$0.45 + x + 0.5 + x = 1 \implies x = 0.025$$ That gives us $P(H_1) = 0.475, P(H_2) = 0.525$ Since there are $10$ arrows and $S_1$ hit target with $8$ this means that: $P(A|H_1) = 0.8$ and same for $S_2$: $P(A|H_2)=0.7$ Now from the law of total probability: $P(A) = P(H_1)P(A|H_1) + P(H_2)P(A|H_2) = 0.7475$ From Bayes' theorem, we have probability that $S_2$ will hit the target: $$P(H_2|A) = \frac{P(H_2)P(H_1)}{P(A)} = \frac{0.525*0.7}{0.7475} \approx 0.492$$ Can someone clarify that I'm not making mistake right now? Last edited by skipjack; May 2nd, 2018 at 03:15 PM. Reason: to correct typos

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