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Thread: Probability Proportion Question

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    Probability Proportion Question




    Hello I am having a bit of difficulty understanding how to go about this question.

    A very large airline claims that 90% of their flight departures are on time.
    If in fact the airline's claim is true, what is the exact probability that in a random sample of 8 flights that there would be as few as 6 departures on time.

    What I did to try and solve this is I multiplied the percent of departures on time for this sample (.75) by the chance that the flights do NOT depart on time (.10). Am I on the right path? Any help is appreciated.

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    Re: Probability Proportion Question


    This is a Binomial Distribution

    {n \choose k}{p^k}{(1-p)}^{n-k}

    where n is the number of trials = 8
    k is the number of successes = 6
    p is the probability of success = .9
    1 - p is the probability of failure = .1

    {8 \choose 6}*{(.9)^6}*{(.1)}^{2}

    = {\ 28}*{0.5314}*{.01}

    \approx 0.1488

    This is probability exactly six flights are delayed.

    If you want the probability 6 or fewer, you could sum P(0), P(1) ... P(6)

    or 1 - ( P(7) + P(8) )

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