At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of $$5$$ phone calls during $$10$$-minute time intervals. The probability that there is at the most one phone call during a $$10$$-minute time period is

The random variable $$X$$ = number of phone calls in a 10-minute interval is given to follow a Poisson distribution with mean (parameter) $$\lambda = 5$$.

For a Poisson distribution the probability of observing exactly $$k$$ events is
$$P(X = k) = \dfrac{\lambda^{k}\,e^{-\lambda}}{k!}$$

The question asks for the probability of “at the most one” phone call, i.e. $$P(X \le 1)$$. Hence

$$P(X \le 1) = P(X = 0) + P(X = 1).$$

Compute each term using the Poisson formula.

For $$k = 0$$:
$$P(X = 0) = \dfrac{5^{0}\,e^{-5}}{0!} = e^{-5}.$$

For $$k = 1$$:
$$P(X = 1) = \dfrac{5^{1}\,e^{-5}}{1!} = 5e^{-5}.$$

Add the two probabilities:

$$P(X \le 1) = e^{-5} + 5e^{-5} = 6e^{-5} = \dfrac{6}{e^{5}}.$$

Therefore the required probability is $$\dfrac{6}{e^{5}}$$.

Option D which is: $$\dfrac{6}{e^{5}}$$