Expected number of shifts before the first malfunction of a poisson process

Question

Consider a poisson process where a machine has $1$ malfunction occur every $3$ hours. Suppose workers work $6$ hour shifts. What is the expected number of shifts that occur prior to to the first malfunction?

Just wanted to make sure I did this correctly or if there is a simpler approach. For a $6$ hour shift, the expected number of malfunctions is $2$ so the probability that no malfunctions happen in a $6$ hour shift is

$$\frac{2^0 \cdot e^{-2}}{0!}=e^{-2}$$

For $X$ to take on the value $0$, we need to have a malfunction in the first shift with probability

$$1-e^{-2}$$

For $X$ to take on the value $1$, we need to have no malfunctions in the first shift and then a malfunction in the second shift with probability

$$(e^{-2})\cdot(1-e^{-2})$$

For $X$ to take on the value $2$, we need to have no malfunctions in the first two shifts and then a malfunction in the third shift with probability

$$(e^{-2})^2\cdot(1-e^{-2})$$

Continuing in this manner, I get an infinite sum in the form

$$\sum_{n=1}^{\infty} n\cdot (e^{-2})^n \cdot (1-e^{-2}) \approx .1565$$

according to Wolfram Alpha.

Answer 1

If you want to model this as a geometric random variable with support $j=0,1,2,...$ and with success probability (chance of at least one malfunction in a shift), $p=1-e^{-2}$ [which comes from $P(X>0)]$, then, as is well documented e.g. on wiki, the expected number of trials before the first success is $qp^{-1}$, where $q=1-p$. This calculates to $\approx 0.1565$, which matches your answer but significantly differs from the expected waiting time to the first arrival in the Poisson model.

However, the Poisson model is more intuitive to me to apply here—if 2 malfunctions happen per shift, on average one occurs by halfway thru, so the mean time (in shifts) until the first malfunction is $1/2$ a shift. In general, the waiting time between arrivals is exponentially distributed with mean $1/\lambda$ where $\lambda$ is the mean rate of arrivals of the process. The time until the first arrival is identical to this waiting time.

By the way, what you calculated was the expectation of a geometric random variable by definition: $\sum_n nq^n p$ and this can be done without wolfram using geometric series and differentiation, might be good practice.

Answer 2

at the outset, your expected number of shifts is E, at the end of the first shift, you either had no incident, so expect a further E shifts - this occurs with probability $e^{-2}$ that you found (at that point your expected shift count is E + 1), or you had an incident so expect zero further shifts, this scenario yields zero shifts and has probability $1 - e^{-2}$

The overall expected number of shifts for scenario 1 & 2 will be

E = P1 x E1 + P2 x E2

so

$E = (1 + E)e^{-2} + 0.(1 - e^{-2})$

$E = (1 + E)e^{-2}$

$E = \frac{e^{-2}}{1 - e^{-2}} \approx .1565$