Example 13. The target consists of 3 pairwise non-intersecting zones. The probability of hitting the first zone with one shot for a given shooter is 0.5. For the second and third zones, this probability is 0.3 and 0.2, respectively. The shooter fires 6 shots at the target. Find the probability that there will be 3 hits in the first zone, 2 hits in the second, and 1 hit in the third zone.
Solution. To find the required probability, we will use formula (3.1.12).

Since in this case $n=6, k_{1}=3, k_{2}=2, k_{3}=1, p_{1}=0.5$, $p_{2}=0.3, p_{3}=0.2$, then

$$
P_{6}(3,2,1)=\frac{6!}{3!2!1!}(0.5)^{3} \cdot(0.3)^{2} \cdot 0.2=60 \cdot 0.125 \cdot 0.09 \cdot 0.2=0.135
$$

Example 14. A treatment method leads to recovery in $90\%$ of cases. What is the probability that out of 5 patients, at least 4 will recover?

Solution. We will use formula (3.1.10), which in this case takes the form $P(A)=P_{5}(4)+P_{5}(5)$, where $A$ denotes the event: "out of 5 patients, at least 4 will recover."

Since

$$
\begin{aligned}
& P_{5}(4)=C_{5}^{4} \cdot 0.9^{4} \cdot 0.1^{1}=5 \cdot 0.6561 \cdot 0.1=0.328 \\
& P_{5}(5)=C_{5}^{5} \cdot 0.9^{5} \cdot(0.1)^{0}=0.9^{5}=0.590
\end{aligned}
$$

To

$$
P(A)=P_{5}(4)+P_{5}(5)=0.328+0.590=0.918
$$