Computers in the lab fail, on average, twice a day, according to a Poisson process. Last week, 10 computers failed. Find the expected time of the last failure, and give an approximate time of day when the last failure occurred.
I've tried a lot.
My first thought was, "ok, arrival times follow a gamma distribution," so I tried to use that and came up with;
$E(S_{10} | N_7 = 10) = \frac{1}{P(N_7 = 10)} \int_0^7 t f_{S_{10}}(t) dt$
Where $S_{10}$ is the time of the tenth arrival and $N_7 = 10$ states that there were 10 arrivals by time 7 (Sunday in this case).
However, the result is extremely large. Should I be using $P(S_{10}<7)$ instead? The probability I have uses the pmf of the poisson distribution. Which, I'm realizing might not be even remotely proper use of math since it's a continuous distribution's expectation divided by a discrete pmf.
Edit: I tried to use $P(S_{10} < 7)$ but it still gets me an incorrect answer. The book claims that the last failure is on Sunday at 8:43 AM. I keep getting Thursday.