Task 3. The average number of airplanes arriving at the airport per minute is 4. Find the probabilities that in 3 minutes a) two airplanes will arrive, b) fewer than two, c) no fewer than two airplanes will arrive.
Solution. We are dealing with the simplest flow, for which $\lambda=4$, so by formula (24) we have: $p_{t}(k)=\frac{(4 t)^{k} \cdot e^{-4 t}}{k!}$.

a) $t=3, k=2, p_{3}(2)=\frac{(4 \cdot 3)^{2} \cdot e^{-12}}{2!}=0.0005$.

b) The event "fewer than two airplanes arrived" is the sum of the events "no airplane arrived" and "one airplane arrived," which are mutually exclusive. By the addition theorem, the required probability is:

$$
p_{3}(k<2)=p_{3}(0)+p_{3}(1)=e^{-12}+12 e^{-12}=13 e^{-12} \approx 0.0001
$$

(by definition $0!=1$).

c) The events "fewer than two airplanes arrived" and "at least two airplanes arrived" are complementary, so $p_{3}(k \geqslant 2)=1-p_{3}(k<2)=$ $=1-13 e^{-12}=0.9999$.

Answer. a) 0.0005; b) 0.0001; c) 0.9999.