TITLE: 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 $P(\text{A hits})$
QUESTION [10 upvotes]: 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?

REPLY [3 votes]: 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 $\frac{(2)}{(2) + (3)}$.