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2 distributions

2 distributions - Probability Distributions Waiting(time of...

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Probability Distributions Waiting (time/# of draws until event) Geometric (discrete) Exponential (continuous) Counting (how many occurrences) – Binominal Poisson (rare events approximation) Normal (Gaussian) -- for continuous values Waiting for an event to occur Let p = probability of an event (say a mutation) occurring on each trail Want to compute the time (or number of trails) until the event – Example: How many individuals do we need to score to find a mutation when the mutation rate is (say) 1/1000? Might expect that this is 1000 individuals, but in reality it’s a bit more complex Probability first n are failures, next is a success is just (1-p) n p The geometric distribution The waiting time (number of trails) until a success is often given by the geometric distribution: Pr(first success on trail n+1) = (1-p) n p What about a success (i.e., AT LEAST one) in the first k? Prob(one, or more, successes in first k) = 1 - Pr(no successes in first k) = 1 - (1-p) k Example: Screening mutations Suppose mutation rate p = 1/1000. How many individuals do we need to score to have a 50% chance of seeing at least one mutation? Solve for k such that 1 - (1-p) k = 0.5 Rearranging gives -(1-p) k = 0.5 -1 = -0.5 Hence (1-p) k = 0.5, or k log(1-p) = log(0.5) Thus k = log(0.5)/log(1-p) = log(0.5)/log(1-1/1000) = 693. Prob(success by 1000) = 1-(1-1/1000) 1000 = 0.63

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Class problems What is prob(at least 1 mutation) for 500 trails?
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