185. The average number of taxi orders received by a dispatch center per minute is three. Find the probability that in 2 minutes: a) four calls will be received; b) fewer than four calls will be received; c) no fewer than four calls will be received.
Solution. According to the condition, $\lambda=3, t=2, k=4$. We will use the Poisson formula

$$
P_{t}(k)=\frac{(\lambda t)^{k} \cdot \mathrm{e}^{-\lambda t}}{k!}
$$

a) The required probability that four calls will be received in two minutes is

$$
P_{2}(4)=\frac{6^{4} \cdot \mathrm{e}^{-6}}{4!}=\frac{1296 \cdot 0.0025}{24}=0.135
$$

b) The event "less than four calls were received" will occur if one of the following mutually exclusive events happens: 1) three calls were received; 2) two calls were received; 3) one call was received; 4) no calls were received. These events are mutually exclusive, so the theorem of addition of probabilities of mutually exclusive events can be applied:

$$
\begin{gathered}
P_{2}(k<4)=P_{2}(3)+P_{2}(2)+P_{2}(1)+P_{2}(0)= \\
=\frac{6^{3} \cdot \mathrm{e}^{-6}}{3!}+\frac{6^{2} \cdot \mathrm{e}^{-6}}{2!}+\frac{6 \cdot \mathrm{e}^{-6}}{1!}+\mathrm{e}^{-6}=\mathrm{e}^{-6}(36+18+6+1)= \\
=0.0025 \cdot 61=0.1525 .
\end{gathered}
$$

c) The events "less than four calls were received" and "at least four calls were received" are complementary, so the required probability that at least four calls will be received in 2 minutes is

$$
P(k \geqslant 4)=1-P(k<4)=1-0.1525=0.8475
$$