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    1) Some of the eggs sold in a store are packed in boxes of 10, for any egg the Probability of it being cracked is 0.05, independent of other eggs. A shelf contains 80 of these boxes. Calculate E(X) of the number of boxes on the shelf that don't contain a cracked egg.

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    I said to myself it's a Binomial distribution and I'd do it for 10 boxes then times it by 8:

    X~B (10,0.95) .
    np= 9.5

    9.5 x 8 = 76 which is wrong.

    2) The random variable X is such that X~B(5,p) , given that P(X=0) = 0.01024, find E(X) and P(X=E(X)) .
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    I'm totally confused on how to find the value of p . If the probability of it not happening at all is 0.01024, then the probability of it happening would be 1-0.01024 right ? So E(X) would be 5xthat, but I end up with a different answer.

    For the 2nd , I'm confused how to find the Probabilty of it , do I do the total number of outcomes divided by the expected value ?
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    1) X~B(80,p) where X=no. of boxes out of 80 with no cracks

    p=prob that a box has no cracks = (0.95)^10.
    Thenuse E(X)=80p
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    2) P(X=0) = 0.01024 means (1-p)^5 =0.01024
    so you can work out p.
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    Thanks for the reply

    For 2 I haven't come across that way before in lessons, is there another way ? And to find the prob of E(X) happening would it be E(X) / 1 ?
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    (Original post by WildBerry)
    Thanks for the reply

    For 2 I haven't come across that way before in lessons, is there another way ? And to find the prob of E(X) happening would it be E(X) / 1 ?
    You should find that p=3/5, so E(X)=5p=3.
    P(X=E(X)) =P(X=3)
 
 
 
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Updated: March 19, 2011
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