The duration of processing a random app in a computer is exponential distributed with parameter $λ = \frac{1}{60}$ seconds. If the computer is processing 20 apps (all apps are similar) in the same time, what is the probability that at least $3$ apps are going to keep being processed for $90$ seconds after they started?
I have thought this:
$$P(X>=90) = 1-e^{-\frac{1}{60}\cdot90} = 0.77$$
and then :
$$P(Y>=3) = 1 - P(Y<3) = 1 - P(Y=0) - P(Y=1) - P(Y=2) = $$ $$=1 - \binom{20}{0}\cdot(0.77)^{0}\cdot(0.23)^{20} - \binom{20}{1}\cdot(0.77)^{1}\cdot(0.23)^{19} - \binom{20}{2}\cdot(0.77)^{2}\cdot(0.23)^{18}$$
I am not sure if this is correct. If not correct, any hints on the problem?
That's almost right, except that $1-e^{-\lambda x}$ is the CDF for being less than or equal to a value. I think you need $e^{-\lambda x}$ for the probability to be greater than a value. Note that whether any particular app runs for 90 seconds is a Bernoulli trial, so you are dealing with a binomial distribution.