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The probability that A hits a target is 1/4 and the probability that B hits a target 1/3. They each fire once at the target.

If the target is hit by only one of them, what is the probability that A hits the target?

I know that this is an independent event. If I do P(A hitting) * P(B not hitting) then (1/4)(2/3) = 1/6 But when I look at the back of my book the answer is 2/5? My book is known to give wrong answers because it is quite old; therefore, I am left with self doubt. Can anyone tell me if I have the correct answer or if I am actually making a mistake?

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    $\begingroup$ This is a conditional probability. Letting $A$ be the event that player $A$ hit the target (in a single shot) and $B$ the event that $B$ hit the target (in a single shot), then what you calculated was $Pr(A\cap B^c)$. What you were told to calculate was $Pr(A\mid (A\cap B^c)\cup (A^c\cap B))$, i.e. the probability that $A$ hit the target given that exactly one of them hit the target. $\endgroup$ – JMoravitz Sep 26 '17 at 4:07
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    $\begingroup$ You might like this explanation of why the formulas being posted here work: arbital.com/p/bayes_rule/?l=1zq $\endgroup$ – Davislor Sep 26 '17 at 14:51
  • $\begingroup$ Would also add the comment that the book is likely not full of mistakes because it is old, but that because it is old the mistakes have been found. New books are not necessarily more correct, they just haven't been around long enough for the mistakes to be as well known. Old does not imply bad. $\endgroup$ – Jared Smith Sep 26 '17 at 15:39
  • $\begingroup$ The probability of A hitting and B not hitting is 1/6 when it is also possible that they both hit or both miss. But in this case, it is not possible for them both to hit or both to miss, so the probability must be greater than 1/6. (If you don't see why, imagine rolling a die. The probability of 1 coming up is 1/6, but only because there are 5 other possibilities. If you roll a die and the result is not a 3, 4, 5, or 6, now the probability it's a 1 is 1/2.) $\endgroup$ – David Schwartz Sep 26 '17 at 18:42
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$$ \begin{align} P(\mbox{target is hit once}) &= P(\mbox{A hitting}) \cdot P(\mbox{B not hitting}) + P(\mbox{A not hitting}) \cdot P(\mbox{B hitting}) \\ &= \frac{1}{4}\cdot\frac{2}{3} + \frac{3}{4}\cdot\frac{1}{3} \\ &= \frac{5}{12} \end{align} $$

So, $$P(\mbox{A hitting | target is hit once}) = \frac{P(\mbox{A hitting}) \cdot P(\mbox{B not hitting})}{P(\mbox{target is hit once})} = \dfrac{\frac{1}{6}}{\frac{5}{12}} = \frac{2}{5}.$$

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    $\begingroup$ I would change the LHS of the last line to "$P(\text{A hitting}|\text{target is hit once})$, to be more explicit that we're using conditional probability here. $\endgroup$ – JiK Sep 26 '17 at 8:18
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Your answer is not correct because you did not account for the case where only B hits, which has probability $\frac13×\frac34=\frac14$. Then the required probability is $$\frac{\frac16}{\frac14+\frac16}=\frac25$$ as the book gives.

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The answer is indeed 2/5 I believe.

\begin{align} \mathbb{P}[\text{A hit | only one hit}] &= \frac{\mathbb{P}[\text{A hit} \,\cap\, \text{only one hit}]}{\mathbb{P}[\text{only one hit}]} \\ &= \frac{\mathbb{P}[\text{A hit}\,\cap\,\text{B didn't hit}]}{\mathbb{P}[\text{A hit}\,\cap\, \text{B didn't hit}] + \mathbb{P}[\text{A didn't hit}\,\cap\, \text{B hit}]} \\ &= \frac{1/4 \cdot 2/3}{1/4 \cdot 2/3 + 3/4 \cdot 1/3} \\ &=\frac{2}{5} \end{align}

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Without using the conditional probability formula:

There are four cases:

  1. Both miss
  2. A hits and B misses
  3. B hits and A misses
  4. Both hit

We're only interested in (2) and (3). (2) has probability $\frac{1}{4}*\frac{2}{3} = \frac{1}{6}$. (3) has probability $\frac{1}{3}*\frac{3}{4}=\frac{1}{4}$. And we need $\frac{(2)}{(2) + (3)}$.

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  • $\begingroup$ What's the last formula if not the conditional probability formula? $\endgroup$ – JiK Sep 26 '17 at 16:48
  • $\begingroup$ @JiK common sense $\endgroup$ – MattPutnam Sep 26 '17 at 17:42
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The probability that only one person hits the target is $$ 1/4 * 2/3 + 1/3 * 3/4 = 5/12 $$ The first event occurs when A hits and B misses, and the second when B hits and A misses. So if only one hit occurs, A hits 2/5 of the time and B 3/5 of the time.

This is an application of Bayes's law. You have a theory: A hit the target. You have data: there's only one hit. What is the probability your theory is true, given the data? 2/5. If you saw two bullet holes, then your theory would be true with probability 1 because A had to hit the target, given those data.

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  • $\begingroup$ We don't consider $P(\text{only one hit}|\text{A hit the target})$ or $P(\text{only one hit}|\text{A didn't hit the target})$ explicitly, so I don't see how this is an application of Bayes's law. $\endgroup$ – JiK Sep 26 '17 at 8:20

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