UE Stochastic Processes - Beispiel 34

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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 \,
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