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How to calculate the expected number of trials before the first success?
It's similar to the binomial distribution, but only considering the first trial.
2 Answers
- siamese_scytheLv 710 years agoFavorite Answer
the probability of getting the first success on the nth trial:
P(n) = p * (1-p)^(n-1)
the mean (expected value) of this distribution is 1/p.
if you index it differently, as n failures and then a success, it becomes P(n) = p * (1-p)^n with mean (1-p)/p.
see geometric distribution:
- cidyahLv 710 years ago
You want the mean (expected value) of the geometric distribution
P(x=1) = p
P(x=2) = (1-p)p
.
.
P(x=k) =(1-p)^(k-1) p
E(x) = 1/p
The first part of the link has the proof.
