Let [tex]X\sim\mathrm{Bin}(400,0.006)[/tex] be the random variable representing the number of genes that do get mutated. Here [tex]\mathrm{Bin}(n,p)[/tex] denotes a binomial distribution with parameters [tex]n[/tex] (total number of genes) and [tex]p[/tex] (probability of mutation).
Then the probability that *at most* 4 genes get mutated is
[tex]\mathbb P(X\le4)=\displaystyle\sum_{x=0}^4\mathbb P(X=x)[/tex]
where
[tex]\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}[/tex]
You should find that
[tex]\mathbb P(X\le4)\approx0.9047[/tex]
