5.3.3 problem 45

Let N be the number of emails received within the first 0.1 hours. Then

$P(T>0.1)=1-P(T<0.1)=1-P(N\ge 3)=1-(1-P(N=0)-P(N=1)-P(N=2\left)\right)=P(N=0)+P(N=1)+P(N=2)$

That is, to find the probability that it takes longer than 0.1 hours for 3 emails to arrive, we find the probability it takes less than 0.1 hours for 3 emails to arrive. To do this, we realize that this is equivalent to at least 3 emails arriving in the first 0.1 hours. And to find that probability, we realize we can find the probabilities of exactly 0, 1, or 2 emails arriving in the first 0.1 hours.

Next, we note that $N\sim \mathrm{Pois}(0.1\ast \lambda )=\mathrm{Pois}\left(2\right)$, per the definition of a Poisson process. Then we get