The probability that A hits a target is $\frac14$ and that of B is $\frac13$. If they fire at once and one hits the target, find $P(\text{A hits})$ 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?       
 A: 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.
A: 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}
A: Without using the conditional probability formula:
There are four cases:


*

*Both miss

*A hits and B misses

*B hits and A misses

*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 $\frac{(2)}{(2) + (3)}$.
A: 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. 
A: $$ \begin{align}
P(\mbox{target is hit once}) &= P(\mbox{A hitting}) \cdot P(\mbox{B not hitting}) + P(\mbox{A not hitting}) \cdot P(\mbox{B hitting}) \\
&= \frac{1}{4}\cdot\frac{2}{3} + \frac{3}{4}\cdot\frac{1}{3} \\
&= \frac{5}{12}
\end{align}
$$
So, $$P(\mbox{A hitting | target is hit once}) = \frac{P(\mbox{A hitting}) \cdot P(\mbox{B not hitting})}{P(\mbox{target is hit once})} = \dfrac{\frac{1}{6}}{\frac{5}{12}} = \frac{2}{5}.$$
