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Math Help - Another probability question

  1. #1
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    Another probability question

    I am having difficulty trying to get correct answer for the following problem:

    Mr. Smith has 8 master keys that unlock several chests. Only one master key will unlock a given chest. 40% of these chests are usually left open, what is the probability that Mr. Smith can unlock a specific chest if he selects 3 master keys at random?

    My attempt: I have a feeling this maybe totally wrong.

    I am pretty sure it is a binomial distribution problem

    P(X=3) = (8C3)(.4^3)(.6^5) = 3.66176
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  2. #2
    MHF Contributor matheagle's Avatar
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    There's a problem just like that in walpole/meyers.
    You need to break it into two cases, the chest being open or locked.
    So it's
    (.4) + (.6)(3/8)
    The .4 is your chance that the chest is open.
    The second pair of numbers is the chance it's locked times the chance
    you selected a key that will open it.
    In walpole they use doors to homes....
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by mathwiz2006 View Post
    I am having difficulty trying to get correct answer for the following problem:

    Mr. Smith has 8 master keys that unlock several chests. Only one master key will unlock a given chest. 40% of these chests are usually left open, what is the probability that Mr. Smith can unlock a specific chest if he selects 3 master keys at random?

    My attempt: I have a feeling this maybe totally wrong.

    I am pretty sure it is a binomial distribution problem

    P(X=3) = (8C3)(.4^3)(.6^5) = 3.66176

    Yikes, a probability larger than 1?
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