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Math Help - Probability question

  1. #1
    Junior Member Misa-Campo's Avatar
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    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?
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  2. #2
    Newbie Coffee Cat's Avatar
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    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.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Misa-Campo View Post
    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
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