finding the probability of not being answered more than two calls!

Question

In a call center , the time between successive calls is exponential random variable with $\mathbf{E}X$= 2 minutes . If the operator is removed for 5 min , what is the probability of not being answered more than two calls\
I found the λ=1/2.And then I tried to find the P(between 0 and 5) but I think this is wrong because says MORE than 2 calls

Answer 1

We can find the parameter of the exponential distribution by $E(X) = 1/\lambda = 2$, so $\lambda = 1/2$. Hence, the arrival process is $Y(t) \sim Poisson(t/2)$, i.e., on average there is one arrival in two minutes. So you are interested in $5$ minutes, i.e., $Poisson(5/2)$, and the event of interest is $\{Y(5)>2\}$, hence \begin{align} P(Y(5)>2) = 1 - P(Y(5)\le 2) &= 1-\sum_{k=0}^2 P(Y(5)=k)\\ &= 1-e^{-5/2} - \frac{5}{2}e^{-5/2} - \frac{e^{-5/2} (5/2)^2}{2!}\\ &= 1-e^{-5/2}\left(1+\frac{5}{2} + \frac{5^2}{2^3}\right) \end{align}