Q : The length of a call is a random variable with an exponential distribution with a parameter (rate). Assume that the above calls are from a factory whose manager decided to interrupt each call that is length exceeds 3 minutes. What is the probability of 100 calls being cut off at most 40 calls (including 40)?
So I calculated the probability of a call over 3 minutes, the exponential distribution, and put an $n$ variable as the exponential distribution rate, cant figure how to solve this question.
Probability for call above 3 minutes:
$ P (x>3) = 1 - P (x<=3) = e^{(-3n)}$ where $n$ is a variable instead of $\lambda$.
then I used binomic distirbution X ~$Bin(100, e^{(-3n)})$
And now stuck.
Let $\lambda^{-1}$ be the mean length of a call, $C$ the threshold for when calls are cut off, $N$ the number of calls made, and $n$ the number of calls in question. The probability that a call exceeds the threshold $C$ is $e^{-\lambda C}$. So the probability of at most $n_0$ calls being cut off is $$ \sum_{n=0}^{n_0} \binom N n e^{-C\lambda n}(1-e^{-C\lambda })^{N-n} $$ Plugging in $C=3$, $N=100$, $n=40$ we have $$ \sum_{n=0}^{40} \binom{100}n e^{-3\lambda n}(1-e^{-3\lambda })^{100-n}, $$ which doesn't really have a nice closed form. For $\lambda=1$ the probability is approximately $1.61247\times 10^{156}$.