# Maths

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
or
= 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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