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Math Help - binomial distrib-how to calculate P(X>3 given n=15, success=.40)?

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    binomial distrib-how to calculate P(X>3 given n=15, success=.40)?

    is it the same as calculating P(X>=3)??
    thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    is it the same as calculating P(X>=3)??
    thanks!
    no. greater than or equal to 3 and greater than 3 are not the same. x > 3 here means x >= 4. it is the same as calculating:

    P(4) + P(5) + ... + P(15)

    but the easier way would be to calculate 1 - P(x <= 3) = 1 - P(0) - P(1) - P(2)
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    how would i do that with the calculator....

    would that be 1-binomcdf(15,.40,2)?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    how would i do that with the calculator....

    would that be 1-binomcdf(15,.40,2)?
    it depends on the calculator. but recall that this is actually what you're doing:

    the probability of k successes in n trials is given by:

    P(k) = {n \choose k}p^kq^{n-k}

    where p is the probability of success, and q = 1 - p is the probability of failure.

    here, p = 0.4, and q = 0.6 and n = 15

    so you want:

    1 - {15 \choose 0}(0.4)^0(0.6)^{15} - {15 \choose 1}(0.4)^1(0.6)^{14} - {15 \choose 2}(0.4)^2(0.6)^{13}


    now, on my calculator, there is a \boxed{_nC_r} button. so to find say, {15 \choose 2} i would enter, "15, \boxed{_nC_r}, 2 = "

    but these are easy enough to work by hand if you're not sure about your calculator.


    {15 \choose 0} = 1

    {15 \choose 1} = \frac {15}1 = 15

    {15 \choose 2} = \frac {15 \cdot 14}{2 \cdot 1} = 15 \cdot 7 = 105
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