Phone calls are received at Janice’s house according to a Poisson Process with parameter $\lambda = 2$ per hour.
How long can Janice’s shower be if she wishes the probability of receiving no phone calls to be at most $0.5$?
My solution : $P(X>t) \leq 0.5$. I got time as $9$ minutes. Is that correct ?
Regarding a Poisson process, you can think of this problem in two different ways: in terms of time, or in terms of counting. Essentially, you're being asked for the first arrival/waiting $W_1$ to be larger than some $t$. Recall that the first waiting time $W_1\sim \text{Exp}(\lambda)$. So you want: $$P(W_1 > t) = e^{-\lambda t} \leq .5$$
However, if I call $N_t$ the number of arrivals up to time $t$, then $$\{W_1 >t\} \iff\{N_t = 0\}.$$ Recall that $N_t\sim \text{Pois}(\lambda t)$. So you want to find: $$P(N_t = 0) = e^{-\lambda t}\frac{(\lambda t)^0}{0!} \leq .5$$ In these two ways, you can solve for $t\leq 20.79442 \text{ min}$. So 20 minutes.