Question

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

a gene will change into mutant gene is 0.006 and is independent of the other genes. What is
the approximate probability that there are at most 4 mutant genes after the treatment?

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`