Question

Suppose a biological cell contains 400 genes. When treated radioactively the probability that

Answer 1

Let $X\sim\mathrm{Bin}(400,0.006)$ be the random variable representing the number of genes that do get mutated. Here $\mathrm{Bin}(n,p)$ denotes a binomial distribution with parameters $n$ (total number of genes) and $p$ (probability of mutation).

Then the probability that *at most* 4 genes get mutated is

$\mathbb P(X\le4)=\displaystyle\sum_{x=0}^4\mathbb P(X=x)$

where

$\mathbb P(X=x)=\begin{cases}\dbinom{400}x0.006^x(1-0.006)^{400-x}&\text{for }x\in\{0,1,2,\ldots,400\}\\\\0&\text{otherwise}\end{cases}$

You should find that

$\mathbb P(X\le4)\approx0.9047$