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Math Help - Expectation Problem

  1. #1
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    Expectation Problem

    Hey guys, homework problem on expectation. This is what i have so far.

    Suppose a fair die is rolled ten times. Find numerical values for the expectations of:

    1) the number of faces which fail to appear in the ten rolls

    Here's what i got:

    E(x) = \sum x P(X = x) = 1 P(X = 1) + 2 P(X = 2) + ... + 5 P(X = 5)

    So the probability that x faces dont is ... this is the part im not sure about.
    P(X = x) =  ?

    I know, for example, that P(X = 5) would have to be (1/6)^10
    After this i am unsure how to proceed. P(X = 1) seems like it would NOT be \left(\begin{array}{cc}10\\5\end{array}\right)(5/6)^{10} ... I cant figure out how to find a general formula for this.

    Thanks in advance!
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  2. #2
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    Response

    I may be missing something, but this is how I see it.

    There are 6 possible faces that may not show, and for each one, the probability is going to be the same (5/6)^10. So for an expected value, I believe it's simply:

    6 * (5/6)^10 = .96903

    Intuitively, this seems like a reasonable answer.
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  3. #3
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    But does that apply when *multiple* faces don't show in ten rolls? Thats what i'm looking for.

    From what i understand, what you've described is the probability a certain face wont show, then multiplied by 6 for each case.
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  4. #4
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    Juicy's answer is correct. Let X = the number of faces not rolled, and let X_i = 1 if face i isn't rolled, 0 otherwise. Then X = X_1 + ... + X_6, and by linearity of the expectation E(X) = E(X_1) + ... + E(X_6) = 6*(5/6)^10.
    Last edited by rn443; October 18th 2009 at 09:15 PM.
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