1. ## Probability question

a) A man fires 5 rounds of bullets at a target, what is the probability that 3 rounds will hit and 2 will miss?

2. 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.

3. Originally Posted by Misa-Campo
a) A man fires 5 rounds of bullets at a target, what is the probability that 3 rounds will hit and 2 will miss?
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:

$P(3)=\frac{5!}{3!(5-3)!}p^3(1-p)^2$

where $p$ is the probability that a single round hits.

RonL