Assume that customers arrive to a shop according to a Poisson process with intensity 2 customers/minute. Let $X(t)$ be the number of customers arriving in a time intervall [0,t].
a) Calculate P$(X(3) = 4)$
I figured I would use the gamma distribution and use $x = 3$.
$$\Gamma(4, 2) = \lambda e^{-\lambda x}\frac{(\lambda x)^{n-1}}{(n-1)!} = 72e^{-6} \approx 0.18$$
However the key says the answer is approximately $0.13$.
On
Reply to OP's query following Graham Kemp's answer (too long for a comment).
There's nothing wrong with the reasoning in your comment per se, except that it's not relevant for the problem you were asked to solve. Your $\ S_n\ $ is the time of the arrival of the $\ n$-th customer. So the cumulative distribution function of your $\ \Gamma(4,2)$-distributed $\ S_4\ $, evaluated at $\ x=3\ $, will tell you the probability that the $4$-th arrival will occur within $3$ minutes. While this is the same as the probability of at least $4$ arrivals occurring during $3$ minutes, what you were asked to find is the probability that that the number of customers arriving in the interval $\ [0,3]\ $ is exactly $4$, which will obviously be smaller.
Also, you appear to have plugged the $\ x=3\ $ into the density function of $\ S_4\ $ rather than its cumulative distribution function, and that won't even give you the correct probability for the $4$-th arrival occurring within $3$ minutes.
Why would you use a gamma distribution?
It is a Poisson process. Use a Poisson distribution.
$$X(t)\sim\mathcal{Pois}(2t)\quad\iff\quad\mathsf P(X(t){=}k)=\dfrac{(2t)^k\mathrm e^{-2t}}{k!}\cdot\mathbf 1_{k\in\Bbb N_0}$$