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# Poisson distribution problems

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A total of M cells are exposed to X-ray radiation. X-rays produce chromosome breakages in cells. For each individual cell i &#1028; {1, .. . , M}, the number N2 of breakages has a Poisson distribution with parameter A (the random variables N are independent). Each breakage has a probability q of healing perfectly, a probability p of healing with a mutation, and a probability r = 1 - p - q of not healing (the associated events are independent between breakages). If all the breakages of a cell are not healed perfectly or healed with a mutation then the cell dies.
a) What is the probability that a given cell dies from the radiation?
b) What is the probability that a given cell survives the radiation without any mutations?
c) What is the probability that a given cell survives the radiation with at least one mutation?
d) What is the average number of mutations per surviving cells?

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#### Solution Preview

The probability that a cell has n breakages is given by the Poisson distribution with parameter lambda:

P(n) = lambda^n/n! exp(-lambda) (1)

To do the problems, it is convenient to use some formal manipulations/techniques. I'll explain these first and give some simple example of these before using them to do the problems. We need to use the series expansion of the exponential function:

exp(x) = sum over k from zero to infinity of x^k/k! (2)

Simple example: We can use (2) to see that the Poisson distribution is properly normalized:

Sum over n from zero to infinity of P(n) =

Sum over n from zero to infinity of lambda^n/n! exp(-lambda) =

exp(-lambda) Sum over n from zero to infinity of lambda^n/n! = (using (2)) =

exp(-lambda) exp(lambda) = 1

To compute averages it is often convenient to modify the probability distribution function by inserting some extra parameters and then do some formal mathematical manipulations on those parameters. Suppose we want to compute the average of n for the Poisson distribution (1). This average is, by definition:

<n> = Sum over n from zero to infinity of n P(n) =

Sum over n from zero to infinity of n lambda^n/n! exp(-lambda)

This summation can certainly be evaluated in a straightforward way. You let the summation run ...

#### Solution Summary

A detailed self contained solution involving power series expansions and generating functions is given.

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