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Math Help - Stuck on simple probability question

  1. #1
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    Question Stuck on simple probability question

    Hello all,

    I've been stuck for the last couple of days on this supposedly very simple probably questions:

    Question:
    30 widgets have been manufactured. 5 of which are known to be faulty. If 5 were randomly selected, what is the probability that exactly 3 of the 5 chosen will be faulty?


    There are a total of 30-choose-5 unique combinations, I've figured out that only 3000 of these combinations contains only 3 widgets that are faulty (aka ignored combinations where 4 and 5 widgets were faulty). This gives a probability of 3000/142506 0r roughly 0.02105

    I was wondering how does one figure this out using probability theory, as the above value of 3000 was found using a computer program that went through all the 30-choose-5 combinations.

    Furthermore I don't think this is a binomial probability problem, as the total number of widgets is known, and the act of replacing does not occur. Assistance will be greatly appreciated.
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  2. #2
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    Quote Originally Posted by Sherrie View Post
    Question:
    30 widgets have been manufactured. 5 of which are known to be faulty. If 5 were randomly selected, what is the probability that exactly 3 of the 5 chosen will be faulty?
    I was wondering how does one figure this out using probability theory, as the above value of 3000 was found using a computer program that went through all the 30-choose-5 combinations.
    This is a simple combination application.
    \dfrac{\binom{5}{3}\binom{25}{2}}{\binom{30}{5}}.

    Where \dbinom{N}{K}=\dfrac{N!}{K!(N-K)!}
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  3. #3
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    Hello Plato,

    Thank-you for the quick response!

    I was wondering if you could explain a bit more about where 25-choose-2 came from and how its being used. I figure the 25 is the result of the remainder of chooseing 5 lots from 30, but in plain terms I don't quite understand how it all comes together.

    For example does the logic change if say 6 widgets were faulty instead of 5, yet 5 are still randomly chosen? would it be (6-choose-3) * (24-choose-2) / (30-choose-5) ?

    btw is there a specific name or term used for the above solution? (its not binomial probability i think.)
    Last edited by Sherrie; April 8th 2011 at 04:38 AM.
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  4. #4
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    Quote Originally Posted by Sherrie View Post
    I was wondering if you could explain a bit more about where 25-choose-2 came from and how its being used. I figure the 25 is the result of the remainder of chooseing 5 lots from 30, but in plain terms I don't quite understand how it all comes together.
    btw is there a specific name or term used for the above solution?
    To get exactly three defective items in a total of five, we pick three from the defective items and two from the twenty-five non-defective items.

    This called sampling without replacement.
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  5. #5
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    Thumbs up

    Quote Originally Posted by Plato View Post
    To get exactly three defective items in a total of five, we pick three from the defective items and two from the twenty-five non-defective items.
    I see, so for every one of the 5-choose-3 combinations there are 25-choose-2 combinations, which explains your original solutions. thanks!
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