Assume arrivals occur according to a Poisson process with average 7 per hour. What’s the probability exactly two customers arrive in the two hour period of time between:
1pm to 2pm and 3 to 4pm
I have this but I'm pretty sure it's incorrect $P(Y=2) = 2\left(\frac{14^2}{2!} \times e^{-14}\right)$
Best Answer
I assume that the following scenarios are counted:
These scenarios are mutually exclusive, so the probability is:
$$\begin{align} \Pr(\text{exactly two})&=\frac{\Gamma^2 e^{-\Gamma}}{2!}\frac{\Gamma^0 e^{-\Gamma}}{0!}+\frac{\Gamma^0 e^{-\Gamma}}{0!}\frac{\Gamma^2 e^{-\Gamma}}{2!}+\frac{\Gamma^1 e^{-\Gamma}}{1!}\frac{\Gamma^1 e^{-\Gamma}}{1!} \\[2ex] &=(\tfrac{1}{2}+\tfrac{1}{2}+1)\Gamma^2 e^{-2\Gamma} \\[2ex] &=2\Gamma^2 e^{-2\Gamma} \\[2ex] &=2\times7^2 e^{-14} \end{align}$$
because there are $\Gamma=7$ expected arrivals per hour.
If my interpretation of the question is correct, your answer is off by a factor of 2.
Since Poissonian events are independent of each other and do not depend on when timing intervals start, the split periods can be treated as a single continuous two hour period, as André Nicolas does.