Sum of iid exponential random variables without using calculus Emails arrive in an inbox according to a Poisson process with rate 20 emails per hour.
Let T be the time at which the 3rd email arrives, measured in hours after a certain fixed
starting time. 
Find $P (T > 0.1)$ without using calculus.
Note that $P (T > 0.1)$ = $1- P (T < 0.1)$.
Let $X_1, X_2, X_3$ be iid $Expo(20)$, then
$T = X_1 + X_2 + X_3$
Clearly, $T$ follows a gamma distribution. with $\lambda = 20$ and $k = 3$.
But in order to evaluate the solution using this way, calculus is needed.
Now the question: How can this question be solved w/o calculus? (e.g., by some kind of story/analogy)
 A: The probability that the waiting time $T$  is $\gt 0.1$ is the probability of $2$ or fewer e-mails in $0.1$ hours. 
This  is given by the probability that a Poisson random variable $Y$, with  parameter $\lambda=(20)(0.1)$, is $\le 2$.
A: Broad Overview:
A Poisson process links two distributions together, the Exponential and Poisson. Recall, the Exponential is often used to model waiting times, while the Poisson is often used to model count data. In terms of the Poisson process specifically, the Exponential models the time until the next arrival and the Poisson models arrival counts (over some time period). This linkage is given the fancy name, the count-time duality.
Poisson process facts:

*

*Arrivals occur at rate $\lambda$. For instance, 3 emails per minute.

*The arrival counts over interval $t = T_{end} - T_{start}$ are distributed $\text{Pois}(\lambda t)$

*Any time interval, $t = T_{end} - T_{start}$, is distributed $\text{Expo}(\lambda)$.

Now, we want to identify two identical events. These two events will serve as our "link." As above, one event will be the waiting time until the third email has arrived at $T_{3}$ since some arbitrary starting point $T_{\text{arb < 3}}$ (recall the "Any" from fact $3$), while the other event will be the email arrival counts prior to $T_3$. To correspond to the problem statement, let's denote this period as $T$, where $T = T_{3} - T_{\text{arb < 3}}$.
If we have to wait at least $.1$ hours until the third email arrives, that means that either $0, 1$ or $2$ emails have arrived in time interval, $T$. Denote the email count over this time period as $N_{t=.1}$. By fact $2$, $N_{t=.1}$ ~ $\text{Pois}(20 \cdot .1) = \text{Pois}(2)$. Then, using our linkage, it follows that, $P(T \gt .1) = P(N_{t=.1} \leq 2) = P(N_{t=.1}=0 \cup N_{t=.1}=1 \cup N_{t=.1}=2) = P(N_{t=.1}=0) + P(N_{t=.1}=1) + P(N_{t=.1}=2) = \frac{e^{-2}2^0}{0!} + \frac{e^{-2}2^1}{1!}  + \frac{e^{-2}2^2}{2!} = 5e^{-2}.$
