Task 1. A radio device contains 1000 elements. The probability of failure of any element within a certain time $T$ is $p=0.004$. Find the probability that exactly 5 elements will fail during time $T$.
Solution. We have $n=1000, p=0.004, q=1-p=0.996, \lambda=$ $=pn=4$. We need to calculate $p_{1000}(5)$. The Poisson formula gives:

$$
p_{1000}(5)=\frac{4^{5} \cdot e^{-4}}{5!}=\frac{1024 \cdot 0.0183}{120}=0.1562
$$

Answer. 0.1562.

Problem 2. The average number of taxi orders per hour is three. Find the probabilities that in two hours there will be: a) 4 orders; b) fewer than 4 orders; c) at least 4 orders.

Solution. We have a simple event flow: $\lambda=3, t=2$, $\lambda t=6$. The Poisson formula takes the form $p_{t}(k)=\frac{6^{k} \cdot e^{-6}}{k!}$. The calculations are as follows:

a) $p_{2}(4)=\frac{6^{4} \cdot e^{-6}}{4!} \approx 54 \cdot 0.00248 \approx 0.134$.

b) $p_{2}(k<4)=p_{2}(0)+p_{2}(1)+p_{2}(2)+p_{2}(3)=e^{-6}+6 e^{-6}+\frac{6^{2} e^{-6}}{2!}+$ $+\frac{6^{3} e^{-6}}{3!}=(1+6+18+36) e^{-6} \approx 61 \cdot 0.00248 \approx 0.1512$.

c) $p_{2}(k \geqslant 4)=1-p_{2}(k<4)=1-0.1512=0.8488$.

Answer. a) 0.134; b) 0.1512; c) 0.8488.