Two dice are thrown simultaneously until two sixes are showing. Find the probability that the process will terminate before the expected number of throws is reached.
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Anonymous2009-01-18 11:36
Sorry, forgot the first part. Basically the text says its geometrically distributed, p = 1/k, E(X) = k, Var(X) = k(k-1)
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Anonymous2009-01-18 14:50
You want cumulative distribution function for up to one less than expected value. First you need probability p of success for a single throw of the dice. That would be 1/36. Anyway, the probability that the kth trial is the first successful trial is p(1-p)^(k-1). The expected value (mean) is k=1/p, or 1/(1/36)=36. The cumulative distribution function for geometric is 1-(1-p)^k = 1-(1-1/36)^36 = 0.63728996689..., but you wanted probability for before the expected number of throws is reached, so really you want the cumulative value for k-1 or 35, 1-(1-1/36)^35 = 0.6269268230... . I've never done this kind before, so take with caution.