the ratio of probability of 3 successes in 5 independent trials to the probability of 2 successes in 5 independent trials is 1/4.
What is the probability of 4 successes in 6 independent trials?

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  1. let prob of success be p
    let prob of failure be q, where p+q = 1

    Prob(3out of 5) = C(5,3) p^3 q^2 = 10p^3 q^2
    prob(2 out of 5) = C(5,2)p^2 q^3 = 10 p^2 q^3

    10p^3q^2/(10p^2q^3 = 1/4
    q = 4p
    in p+q=1
    p + 4p=1
    p = .2 , then p = .8

    so prob(4 out of 6) = C(6,4) (.8)^4 (.2)^2
    = .24576
    = 768/3125

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  2. p^3(1-p)^2*[5!/(2!*3!)] = P(3 out of 5)
    p^2*(1-p)^3*[5!/(3!*2!)]= P(2 out of 5)

    p/(1-p) = 1/4
    p = 1/5 is the single-trial probability of success

    P(4 out of 6) = p^5(1-p)^2*[6!/(4!*2!)]
    = 3.2^10^-4*(0.64)*15 = 0.0307

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  3. looks like I did not mind my p's and q's

    from p+4p=1
    p=.2, then q= .8

    prob(4 out of 6 successes) = C(6,4) (.2)^4 (.8)^2
    = .01536

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