Exponential problem - Probability that we have to wait more than 4 min at least 3 times

Question

Suppose the time it takes to reach a representative when calling customer service is exponentially distributed with a mean of 2.5 minutes. If we call the customer service 11 times in a month without ever hanging up, what is the probability that we have to wait more than 4 minutes at least 3 three times?

I know that the $\lambda = 1/2.5$ for the exponential distribution when we call once so when we call them 11 times is $\lambda = 11*(1/2.5)$? I'm a bit confused, can anyone explain this a bit?

Answer 1

The frequency distribution and CDF for an exponential function:

$f(x) = \frac 1{\beta} e^{-\frac x\beta}\\ F(x) = \int_0^{x} f(x)\ dx = 1-e^{-\frac x\beta}$

The probability that any one call takes longer than 4 minutes.

$e^{-\frac {4}{2.5}}$

At least 3 is not 0, 1 or 2

The probability that no calls take more than 4 minutes

$(1-e^{-\frac {8}{5}})^{11}$

exactly one call takes more than 4 min

$11(1-e^{-\frac {8}{5}})^{10}(e^{-\frac {8}{5}})$

2 calls

$55(1-e^{-\frac {8}{5}})^{9}(e^{-\frac {8}{5}})^2$

$1-(1-e^{-\frac {8}{5}})^{11} - 11(1-e^{-\frac {8}{5}})^{10}(e^{-\frac {8}{5}}) - 55(1-e^{-\frac {8}{5}})^{9}(e^{-\frac {8}{5}})^2$

Answer 2

Outline:

On any one given call, what is the probability that you have to wait more than $4$ minutes? Call this probability $p$. You should be able to calculate this probability from the exponential distribution.

You are repeating this process $11$ times (presumably independently). The distribution of the number of times you have to wait more than $4$ minutes is therefore binomial($n=11$, $p$).