A particle detector has a flaw that makes it get stuck for 10 seconds each time it detects a particle. When it is not stuck assume 100% efficiency (it detects every particle).
Given that particles that pass through the detector follow a Poisson distribution that in 100 seconds got 38.4 particles on average, determine how many times the detector will get stuck in an hour.
It is easy to see that on average we expect
$$ 38.4 particles / 100 seconds = 0.384 particles / second$$
Also it is easy to know that with a number of particles very big, the detector will get stuck 3600 seconds / 10 seconds/times = 360 times.
I can predict how many particles will pass through in a given hour, we just multiply the rate by 3600 seconds in an hour but how to account for the percentage of times in which a particle comes when the detector is already stuck so it cannot be stuck a second time?
The amount of times it gets stuck surely must be less than the number of particles going through but how much quantitatively?
By linearity of expectation, the expected number of times the detector gets stuck is the expected number of particles that arrive times the probability that the detector isn't stuck at any given time.
When a particle arrives, the detector gets stuck for $10$ seconds, and then on average it remains unstuck for $0.384^{-1}\mathrm s$. Thus the proportion of time it isn't stuck is
$$ \frac{0.384^{-1}}{10 + 0.384^{-1}}=\frac1{3.84+1}=\frac1{4.84}\;, $$
and the expected number of times it gets stuck in an hour is
$$ 3600\cdot0.384\cdot\frac1{4.84}\approx286\;. $$