Poisson distribution question about expectation

Question

Solving this exercise right now:

New files arrive at an office according to a Poisson process with a rate of $5$ per hour.

a) What is the probability the office receives exactly two files between $2$ and $2$:$30$?

We record the arrival of the first five files on a certain day.

b) What is the probability that between every two consecutive files recorded, there's a gap of at least $12$ minutes?

c) What is the expected duration in minutes between the $2$nd and $4$th files?

I solved a) and b). For a), I simply did $P(X=2)$ with the rate being $2.5$ instead, and I got $0.257$. For b), I changed the rate to $1$ since that's the rate in intervals of $12$ minutes, and then I got $[P(X=0)]^4$, which is $0.0183$.

My problem is with c). Any idea on how to do that part?

Answer 1

The time between two jumps of a Poisson process has an exponential distribution. Since the Poisson process has independent increments then it will be just $2\,\mathbb{E}X=1/\lambda$

Answer 2

For part (c), the expected time from the 2nd to the 4th file is twice the expected time from the 2nd to the 3rd, so notice that \begin{align} & \Pr(\text{time from 2nd the 3rd} > t) \\[6pt] = {} & \Pr\big((\text{number of arrivals during time $t$}) =0 \big) \\[6pt] = {} & \frac{(5t)^0 e^{-5t}}{0!} = e^{-5t}. \end{align} Therefore, the time from the 2nd to the 3rd has an exponential distribution with rate $5\text{ per hour.}$ (The answer by Pedro Ignacio Martinez Bruera asserts that but does not explain why that must be so.) That has expected value $(1/5) \text{ hour}.$