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binomial probabbility

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Assume that a procedure yields a binomial distribution with a trial repeated n times. use the binomial probability formula to find the probability of x success given the probability p of success on a single trial.

1. by formula- n=9, x=2, p=0.35

2. by PDF- n=15, x=13, p=1/3

  • binomial probabbility -

    C(9,2) = 9!/[2!(7!)]
    = 9*8/2 = 9*4 = 36

    so 36(.35)^2(1-.35)^7
    = 36 * .1225 * .049
    = .216

    I do not know why by PDF means (By probability distribution function?)

  • binomial probabbility -

    yes thank you I believe that's what it stands for...

  • binomial probabbility -

    Well, I suppose I could look it up in a table of the distribution for n = 15, x =13, p =.3333... by symmetry
    it is the same as for n = 15, x = 2, p = .33333....
    but I do not have a good table
    I have only up to n = 10
    so have o do it the same old way
    C(15,13) = 105
    so
    105 * (1/3)^13 * (2/3)^2
    = 2.927*10^-5
    c(15,13)

  • binomial probabbility -

    Now maybe they mean using the normal function which is the limit of binomial for large n

    u = n p = 15*1/3 = 5
    s^2 = 5(2/3) = 10/3 = 3.333...
    s = 1.83

    z = (x-u)/s = (13-5)/1.83 = 4.38
    f(z) = (1/sqrt(2pi) e^-(4.38^2/2)
    = 2.72 * 10^-5
    gee, not too far off

  • binomial probabbility -

    For a binomial distribution with parameters
    n = 5 , p = 0.3 . Find the probabilities of
    getting :
    (i) Atleast 3 successes.
    (ii) Atmost 3 successes.
    n = 5 , p = 0.3

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