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May 2nd, 2018, 02:59 AM   #1
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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.
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May 2nd, 2018, 10:01 AM   #2
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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?
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May 2nd, 2018, 01:03 PM   #3
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$A$ - event that archers will hit target
$H_1$ - 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
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