a) A man fires 5 rounds of bullets at a target, what is the probability that 3 rounds will hit and 2 will miss?
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.
Could be anything without more information. If you specify the hit probability for a single round and specify that all rounds have the same chance of hitting and are independent. Then the number of hits has a binomial distribution so the probability of 3 hits from 5 rounds is:
$\displaystyle P(3)=\frac{5!}{3!(5-3)!}p^3(1-p)^2$
where $\displaystyle p$ is the probability that a single round hits.
RonL