# Probability Of Queueing (Internetworking)

Suppose

• Total Users: $n$
• Each has a probability of $p$ being active

The probability that exactly $k$ users are present is:

$\binom{n}{k} \times (1-p) ^ {n-k} \times p^k$

For example, for a population of 35 independent users, each with a probability 0.1 of being present, the likelihood that > 10 users are current is:

$\sum_{k=10}^{n=35} \binom{n}{k} \times (1-p) ^ {n-k} \times p^k$