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 May 2nd, 2018, 02:59 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 03:02 AM. May 2nd, 2018, 10:01 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 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, 01:03 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, \begin{equation} P(H_1) = \frac{\frac{9}{10}}{2} + x = 0.45 + x \end{equation} and same for other archer: \begin{equation} P(H_2) = \frac{\frac{10}{10}}{2} + x = 0.5 + x \end{equation} Now we have: \begin{equation} 0.45 + x + 0.5 + x = 1 \implies x = 0.025 \end{equation} 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: \begin{equation} P(H_2|A) = \frac{P(H_2)P(H_1)}{P(A)} = \frac{0.525*0.7}{0.7475} \approx 0.492 \end{equation} Can someone clarify that I'm not making mistake right now? Last edited by skipjack; May 2nd, 2018 at 02:15 PM. Reason: to correct typos Tags bayes, problem, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post amphinomos Probability and Statistics 2 January 20th, 2015 11:43 PM PinkMarill Probability and Statistics 10 July 16th, 2014 03:17 PM Damoo Algebra 1 November 21st, 2012 04:21 PM FreaKariDunk Algebra 2 September 21st, 2012 08:23 AM hejmatematik Algebra 2 December 8th, 2011 11:00 PM

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