# Probability of genetic experiment on cell division

• Jun 1st 2010, 06:47 AM
acevipa
Probability of genetic experiment on cell division
A genetic experiment on cell division can give rise to at most 2n cells. The probability distribution of the number of cells X recorded is:

$\displaystyle P(X=k)=\frac{\theta^k(1-\theta)}{1-\theta^{2n+1}}$ $\displaystyle for\ 0 \leq k \leq 2n$

where $\displaystyle \theta$ is a constant with $\displaystyle 0<\theta<1$

What are the probabilities that:

1) An odd number of cells is recorded?

2) At most n cells are recorded?
• Jun 1st 2010, 06:09 PM
ANDS!
Without actually values for $\displaystyle \theta$ and N, you're not going to get very far. However:

A. You're literally just plugging in the values you are given. K is dependent on the number of cells before the experiment is conducted, so K can be an odd or an even number: 2N-1 (odd) or 2N (even).

B. Same rationale as above, but you're going to have to use some reasoning. If "n" cells are recorded after the experiment is conducted, how many cells were present before (the answer depends on whether n is even or odd).

Assuming that's ALL the information you are given for the problem, I don't see much more you can do with this.