# UE Stochastic Processes - Beispiel 34

In an office telephone messages arrive according to a Poisson process at the mean rate of six per hour and fax messages at three per hour.

$X \sim \mbox{Poisson}(6, t), Y \sim \mbox{Poisson}(3, t) \,$


(a) Find the probability that exactly two messages (phone or fax) are received between 9.00 and 9.40.

$40 \mbox{ minutes} = 2/3 \mbox{ hours} \Rightarrow t = 2/3 \,$

Suchen $P(X+Y = 2)\,$

$X+Y \sim \mbox{Poisson}((6+3),t) \,$

$P(X+Y = 2) = 0.0446\,$


(b) Find the probability that the first message after 10.00 occurs before 10.10.

$10 \mbox{ minutes} = 1/6 \mbox{ hours} \Rightarrow t = 1/6\,$

$Z = X+Y\,$

$P(Z \ge 1) = 1 - P(Z < 0) = 1 - P(Z = 0) = 0.7769 \,$