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