# UE Stochastic Processes - Beispiel 35

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.?

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


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polyroot always returns complex

erg=Re(polyroot(c(-0.1,1,1,-1)))
erg
log(erg)/-3


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