Question

A population of cells has been exposed to an experimental drug G resulting in 6% of cells becoming senescent. Consider an experiment in which you sequentially test individual cells (for example using a flow cytometer) to determine whether they are senescent.

a) What do you need to assume for a geometric probabilistic model to apply? 


b) Using a geometric probabilistic model, determine if you run the experiment multiple times (on 
the same cell population), on average, how many cells you must examine until you find a 
senescent cell? What is the variance? 


c) For the experiment above, what is the probability the first cell you test is senescent? Interpret 
the result. 


d) For the experiment in (b), what is the probability you have to examine five or more cells before you encounter the first senescent cell? 


Answer 1

Let X be no. individual cells examined until to find a senescent cell.

$Then~X\sim Geometric(p)~\\ where~p=probability~of~getting~senescent~cell=0.06\\ The~pmf~of~X~is\\ P(X=x)=q^{x-1}p,~~x=1,2,....;~q=1-p\\ =0~otherwise\\ a.~E(X)=\sum_{x=1}^\infty xq^{x-1}p=p\sum_{x=1}^\infty \frac{\mathrm{d} q^x}{\mathrm{d} q}=p\frac{\mathrm{d} }{\mathrm{d} q}\left(\sum_{x=1}^\infty q^x \right )=p\frac{\mathrm{d} }{\mathrm{d} q}\left(\frac{q}{1-q} \right )\\ =p\times \frac{1-q+q}{(1-q)^2}=\frac{1}{p}=16.67\approx 17\\ E(X(X-1))=\sum_{x=1}^\infty x(x-1)q^{x-1}p=p\sum_{x=1}^\infty \frac{\mathrm{d} q^x}{\mathrm{d} q}=pq\frac{\mathrm{d}^2 }{\mathrm{d} q^2}\left(\sum_{x=1}^\infty q^x \right )=pq\frac{\mathrm{d}^2 }{\mathrm{d} q^2}\left(\frac{q}{1-q} \right )\\ =pq\times \frac{2}{(1-q)^3}=\frac{2q}{p^2}\\ Var(X)=E(X(X-1))+E(X)-E^2(X)=\frac{q}{p^2}\\ =(1-0.06)/(0.06)^2= 261.1111\\ c.~P(X=1)=p=0.06\\ i.e.~6\%~chance~that~the~tested~first~cell~is~senescent.\\ d.~P(X\geq 5)=\sum_{x=5}^\infty q^{x-1}p=pq^4\sum_{x=5}^\infty q^{x-5}=pq^4\times \frac{1}{1-q}=p^4\\=(0.06)^4=0.000013$