How many cells must I sample to have at least 3 mutated cells? I am attempting to design an experiment where I am looking for rare mutated cells in a large cell population.  The cell population is ~100 million cells and about ~100 thousand of these cells are mutated.  I am trying to figure out how to calculate how many cells I have to sample (without replacement in the population) to be at least x% (say 90%, 95%, 99%) sure that I have at least 3 cells sampled that have the mutation.
 A: In the first place, with a cell populations of 100 million, I wouldn't worry so much about sampling with/without replacement.  I'd be much more interested in whether your population is well mixed so that each sample could be considered to have the same chance of being a mutated cell (i.e. 1 in 1000)
If that's true then the number of mutant cells in a sample of size n is going to follow a Binomial distribution.  There's a handy binomial calculator [here] (http://stattrek.com/online-calculator/binomial.aspx)
5500 samples would get you a probability of 90.5% that you'd have 3 or more mutant cells.  7000 samples will get you a probability of 95.8% that you'd have 3 or more mutants. You might want to quickly explore how sensitive your sample size needs to be for different probabilities.  For events this rare, small changes can make very large differences in necessary sample sizes.
If you need the mathematics behind the calculations I'd be happy to help with that as well, but it sounds just like you're wanting to run an experiment and not explore the math.
A: Let $M_n$ be the number of mutated cells found in a sample of $n$ cells.
Let $p$ be the probability that a cell is mutated.  The population is large enough for this to be assumed independent and identically distributed.  This is $p=\frac{10^5}{10^6}= \frac 1{10}$
Thus $P(M_n=x) = {n \choose x} p^x (1-p)^{n-x}$
This is a binomial distribution.  $M_n \sim \mathcal{B}(n, p)$
$\begin{align}P(M-n\geq 3) & = 1 - (1-p)^n - np(1-p)^{n-1} - (n-1)np^2(1-p)^{n-2} \\ & = 1 - (0.90)^n - n (0.10)(0.90)^{n-1} - (n-1)n(0.010)(0.90)^{n-2} \end{align}$
Either solve this for $n$ by whatever calculator you have at hand (Matlab, etc), or use an approximation.

This can be approximated by the normal distribution: $M_n \dot\sim \mathcal{N}(np, np(1-p))$
Or when normalised  $\dfrac{M_n-np}{np(1-p)}\dot\sim\mathcal{N}(0,1)$ 
So look up the relevant $\alpha$ values for the standard normal distribution and solve 
Find $\alpha$ where $P(Z\leq \alpha) = 0.90$, etcetera, and solve $\frac{3-np}{np(1-p)}=\alpha $ for $n$.
