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
(1p)
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) = (1p)
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  (1p)
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  (1p)
k
= 0.5
–
Rearranging gives (1p)
k
= 0.5 1 = 0.5
–
Hence (1p)
k
= 0.5, or k log(1p) = log(0.5)
–
Thus
k = log(0.5)/log(1p)
–
= log(0.5)/log(11/1000) = 693.
–
Prob(success by 1000) = 1(11/1000)
1000
= 0.63
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Class problems
•
What is prob(at least 1 mutation) for 500
trails?
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 Spring '07
 WEINERT
 Normal Distribution, Probability theory, 50%, 90%, 98.5%, 11 %

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