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Thread: Binomial Probablity Distribution exactly one & and at least one.. check my answer

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    Unhappy Binomial Probablity Distribution exactly one & and at least one.. check my answer




    Your favorite baseball player has a 0.300 batting average (that is, he averages three hits for every 10 times he is up to bat). If he is up to bat five times in a particular game, what is the probability that he will get exactly 1 hit and at least 1 hit?
    P(X) = n! / x! (n-x)! p^x (1-p)^(n-x)

    P(X) = 5! / 1! (5-1)! (.30)^1 (1-.30)^5-1

    P(x) = 120 / 1 (24) (.30) (.24)

    P(X) = 5 (.30) (.24)

    P(X) = Exactly once = .36
    P(X) = At least once = 1 -.36 = .64

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    Re: Binomial Probablity Distribution exactly one & and at least one.. check my answer

    The exactly once answer is right but you might want to rethink your "at least once" approach.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Binomial Probablity Distribution exactly one & and at least one.. check my answer

    Thanks so much !! I take one right out of two !! Let me rework !

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    Re: Binomial Probablity Distribution exactly one & and at least one.. check my answer


    I looked at my notes and it said the following "at least x" occurrences can be found from subtracting the appropriate cumulative probability from 1 using the Cumulative binomial probability distribution table

    We can calculate P(X ≥ 1) by finding P(X ≤ 0) and subtracting it from 1

    So I went to the table
    P = .30
    n= 5
    x success 0 1-0.168 = 0.832

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