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# of bernouli trials to get one success
p.m.f= (1-p)^(k-1)*(p)
var=(1-p)/(p^2) |
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p=prob in n=trials, usually finding probability of getting some X<=n p.m.f= (n choose k)(p^k)(1-p)^(n-k)
var=np(1-p) |
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given lambda
expected value=lambda L
var=lambda
P(X=k)=(e^-L)((L^k)/(k!)) |
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| P(X|Y)=P(X given Y)=((P(Y|X)*P(X))/(P(Y|X)*P(X)-P(Y|Xc)*P(Xc)) |
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(1-p)^n
this is obvious though. think about it. the first success has to be greater than n if and only if there are at least n failures in a row first. |
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