Problem 5. The probability of at least one hit with two shots is 0.96. Find the probability of four hits with five shots.
Solution. By the Bernoulli formula (19), we have $p_{5}(4)=C_{5}^{4} \cdot p^{4} \cdot q^{1}$. Here, the probability $p$ of hitting the target with one shot is unknown. To find it, we use the condition of the problem.

Let's denote the events:

$A_{1}$ - hitting the target with the first shot;

$A_{2}$ - hitting the target with the second shot;

$A_{1}+A_{2}$ - hitting the target at least once in two shots.

By formula (12), we find $p\left(A_{1}+A_{2}\right)=1-q^{2}$, i.e., $0.96=1-q^{2}$, from which $q=0.2$ and $p=1-q=0.8$. Therefore, $p_{5}(4)=5 \cdot(0.8)^{4} \cdot 0.2=$ $=0.4096$.

Answer. 0.4096.