I believe this problem is missing a few bits of data. The probability that a bullet will hit a target is not given. Assuming that the probability of a hit and the probability of a miss are equal (which is very unlikely in reality, since it's usually easier to miss):

probability of hit: 1/2

probability of miss: 1/2

There are many ways wherein you can have 3 hits and 2 misses. One situation is:

1st: hit

2nd: hit

3rd: hit

4th: miss

5th: miss

The probability of this specific occurrence is:

1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32

Now, this isn't the only way to have 3 hits and 2 misses. The number of ways to accomplish this is:

5C3 2C2 = 10

That's combination of 5 taken 3 at a time (take 3 out of the 5 shots and make them hit), multiplied by combination of 2 taken 2 at a time (take 2 out of the 2 remaining shots and make them miss).

So the total probability is 10/32.