# Bacterial Plates Mutations

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• Feb 12th 2010, 03:32 PM
sirellwood
Bacterial Plates Mutations
Bacterial plates are exposed to radiation. The number of mutations caused by the radiation has a Poisson distribution wth mean $\displaystyle \mu$. We can only observe if a plate has no mutations or at least one. N plates are examined. The number of mutations on each plate can be assumed to be independent of each other.

Let Yi = 0 if the ith plate has no mutations and Yi = 1 if the ith plate has at

least one mutation. Z = $\displaystyle \Sigma$ Yi, i=1....N, is the number of plates with at least one mutation, derive E(Z).
• Feb 13th 2010, 12:04 PM
eigenvex
Since the Yi's can only take on values of 0 or 1 with a set probability, they are Bernoulli trials. You can find the probability of having NO mutations on a plate by using the poisson distribution for k = 0 (k = # of occurances). Doing 1 - this probability gives you the probability of having at least 1 mutation on a plate. With this probability, you get your parameter for the Bernoulli RV's, and so you have the distribution of your Yi's.

Then, to find E(Z), just take the expectation inside of the summation (since expectations are linear operators). The Yi's are independent so you can find the expected value of each one and then sum them up from 1 to N to get the total expected value, E(Z).
• Feb 15th 2010, 02:45 AM
sirellwood
Ok, im struggling a bit.... so far iv got....

P(0 successes) = e^-mu

and P(>0 successes) = 1 - (e^-mu)

i dont really understand where im meant to go from here, or if even that is the right start!
• Feb 15th 2010, 06:40 AM
eigenvex
Ya, that's correct. Now you know the probabilities of the Yi's having successes (a mutation) or a failure (no mutation).

Expected values are the outcomes weighed by their probabilities:

$\displaystyle E(Yi) = (1)(1-e^{-\mu}) + (0)(e^{-\mu})$

and Z is the sum of all the Yi's, which all have the same expectation, so:

$\displaystyle E(Z) = E(\Sigma Yi) = \Sigma(E(Yi)) = NE(Yi)$