UE Stochastic Processes - Beispiel 35

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Customers arrive at a bank according to a Poisson process at a mean rate of \lambda = 10\, per minute. A proportion 0.6 wish to draw out money (type A), 0.3 wish to pay in money (type B) and 0.1 wish to do something else (type C). 

Wegen UE Stochastic Processes - Beispiel 31 

A \sim \mbox{Poisson}(6, t) \,

B \sim \mbox{Poisson}(3, t)\,

C \sim \mbox{Poisson}(1, t)\,

 P(N(t) = k) = e^{-\lambda t} \frac{(\lambda t)^{k}}{k!} \,

(a) If twenty customers arrive in two minutes, what is the probability that just one is of type C? 

\begin{align}
 P(C=1|A+B+C = 20) 
& = \frac{P(C=1 \cap A+B+C = 20)}{P(A+B+C = 20)} \\
& = \frac{P(C=1) \cdot P(A+B = 19)}{P(A+B+C = 20)} \\
& = 0.2702
\end{align}

(b) How much time must elapse before at least one customer each of type A and B will have arrived with a probability of 0.9.? 

\begin{array}{rcl}
P(A \ge 1, B \ge 1; t) & = & 0.9 \\
P(A \ge 1;t) \cdot P(B \ge 1; t) & = & 0.9 \\
(1 - P(A = 0;t)) \cdot (1 - P(B = 0; t)) & = & 0.9 \\
\left(1 - e^{-6 t} \frac{(6 t)^{0}}{0!} \right) \left(1 - e^{-3 t} \frac{(3 t)^{0}}{0!} \right) & = & 0.9 \\
(1 - e^{-6 t}) (1 - e^{-3 t}) & = & 0.9 \\
1 - e^{-3 t} - e^{-6 t} + e^{-9 t} & = & 0.9 \\
e^{-3 t} + e^{-6 t} - e^{-9 t} & = & 0.1 \qquad x := e^{-3t}\\
 - 0.1 + x + x^2 - x^3& = & 0 \\
\end{array}


# polyroot always returns complex
erg=Re(polyroot(c(-0.1,1,1,-1)))

erg

log(erg)/-3

#
[1]  0.09227156 -0.68180750  1.58953594

[1]  0.7943398        NaN -0.1544807



Weil t = - \frac{log(x)}{3}\, und t > 0\,, muss t = 0.7943398\, sein.
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