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Thread: Bionomial Probability Distributions

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    Bionomial Probability Distributions





    51% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan. Find the probability that the number who consider themselves baseball fans is (a) exactly five, (b) at least six, and (c) less than four

    (a) P(5) = 10C5 •(0.51)^5 • (0.49)^10-5







    I tried putting this into the calculator but it was wrong. I know that 10 is the number of trials and 51% is p. I looked at a couple of examples online but I'm still a little lost. I set up the equation/problem above but how do you go about solving it?
    Last edited by rihnavy; 09-30-2016 at 05:32 PM.

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    Re: Bionomial Probability Distributions

    The formula is correct.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: Bionomial Probability Distributions

    Is it the value of 10c5 that is causing you problems? Try 10!/(10-5)!5! which is 252

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    Re: Bionomial Probability Distributions

    Quote Originally Posted by CE479 View Post
    Is it the value of 10c5 that is causing you problems? Try 10!/(10-5)!5! which is 252
    Thank you! I solved a but now how do I go about solving for B and C? The equation for b is 10C6(0.51)^6(0.49)^4 right?

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    Re: Bionomial Probability Distributions

    Thank you! I solved a but now how do I go about solving for B and C? The equation for b is 10C6(0.51)^6(0.49)^4 right?

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    Re: Bionomial Probability Distributions


    If it's at least 6, then you have to add the probabilities for 6,7,8,9 and 10 together. Similar reasoning for c.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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