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Thread: Binomial Distribution confusion

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    Binomial Distribution confusion




    Hi, I'm currently confused on a question and don't quite understand how to solve it.

    The question is Suppose 75% of all households with telephone service have voicemail. You are conducting a survey about voicemail usage and you randomly call 10 households.

    What is the probability that the first 7 households have voicemail and the last 3 do not?


    So, how exactly would I do both the first 7 households have voicemail and the last 3 don't? I've tried doing P(X<=7) but not sure how to incorporate the "3 do not" part.

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    Re: Binomial Distribution confusion


    hi,
    it is not a binomial distribution stricly. You need to calculate P(A*B) where A is getting 7 voicemails out of 7 and B is getting 3 without a voicemail out of 3.

    Regards

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