*Poisson distribution* is a "discrete probability distribution. It expresses the probability of a number of events occurring in a fixed time if these events occur with a known average rate, and are independent of the time since the last event" [Wiki]. Such events are said to be memoryless.

Most queuing systems' characteristics such as arrival and departure processes are described by a poisson distributions. Assuming that arrivals and departures are random and independent i.e. they exhibit pure-chance property; arrivals are described by a poisson random variable or poisson random distribution as shown by equation 1.1 [Wiki, Char]

The probability that there are exactly k occurrences (k being a non-negative integer, k = 0, 1, 2, ...) is

Where

e is the base of the natural logarithm (e = 2.71828...),

k! is the factorial of k,

λ is a positive real number, equal to the expected number of occurrences that occur during the given interval. For instance, if the events occur on average every 4 minutes, and you are interested in the number of events occurring in a 10 minute interval, you would use as model a Poisson distribution with λ = 2.5.

(taken from Bajpai A.C, 1974)
The average rate of telephone calls received at an exchange of 8 lines is 6 per minute. Find the probability that a caller is unable to make a connection if this is defined to occur when all lines are engaged within a minute of the time of the call.

We first need to make an assumption that the overall rate of calls is constant, then we can use equation 1.1 as follows:
Since our time unit is 1 minute, then λ = 6

Which leads to

The probability of not being able to make a call occurs only when there are at least 9 calls in any interval of a minute.

So, this leds to p(k+1) equals to:

Solving this bit by bit with k from 0 to 8, we get p(k+1) = 0.8546.