103. Within an hour, a switchboard installed to connect telephone extensions in the offices of a trading company receives on average 90 calls. Assuming that the number of calls on any time interval follows a Poisson distribution,

a) construct the distribution series of the random variable $X$ - the number of calls received by the switchboard in 4 minutes;

b) find the probability that at least three calls will be received in 4 minutes.
Solution. The random variable $X$ - the number of calls received by the switchboard in four minutes, is distributed according to the Poisson law. On average, 90 calls are received by the switchboard per hour, so the average number of calls received by the switchboard in four minutes is $\frac{90}{60} \cdot 4=6$. This means that the expected value of the number of calls in 4 minutes is $M(X)=\lambda=6$.

a) Using the Poisson formula (21), we find the probabilities:

$$
P_{0}=\frac{\lambda^{0}}{0!} e^{-6}=1 \cdot e^{-6} \approx 0.0025
$$

$$
\begin{aligned}
& P_{1}=\frac{\lambda^{1}}{1!} e^{-6}=6 \cdot e^{-6} \approx 0.0149 \\
& P_{2}=\frac{\lambda^{2}}{2!} e^{-6}=\frac{6^{2}}{2} e^{-6} \approx 0.0446 \\
& P_{3}=\frac{\lambda^{3}}{3!} e^{-6}=\frac{6^{3}}{6} e^{-6} \approx 0.0892
\end{aligned}
$$

and so on. These same values of $P_{m}$ for $\lambda=6$ can also be found in the Poisson distribution table (see table 2 in the appendix).

Using the found values of $P_{m}$, we construct the distribution series of the random variable $X$:

$$
X: \begin{array}{cccccccc}
x_{i} & 0 & 1 & 2 & 3 & \cdots & m & \cdots \\
p_{i} & 0.0025 & 0.0149 & 0.0446 & 0.0892 & \ldots & \cdot \frac{\lambda^{m}}{m!} e^{-\lambda} & \cdots
\end{array}
$$

b) Consider the event $B$ - the receipt of at least three calls by the switchboard in four minutes. The opposite event to this is $\bar{B}$ - the receipt of fewer than three calls by the switchboard in four minutes.

The probability of event $B$ is $P(B)=1-P(\bar{B})$, i.e.,

$$
P(X \geq 3)=1-P(X<3)=1-(P(0)+P(1)+P(2)) \approx 1-(0.0025+0.0149+0.0446)= \\
& =1-0.062=0.9380 .
\end{aligned}
$$