inter-arrival time between customers to a restaurant with an exponential distribution The interarrival time between customers to a restaurant is given to follow Exp(0.1) in minutes, (including the time between their opening and arrival of the 1st customer)  what is the probability that the first 5 customers arrive within 40 min?
I come with this result:
$\lambda = 40*0.1$
$\Pr(X=k) = (\lambda^k*e^{-\lambda})/k!$ 

$((0.1*40)^5*\exp(-0.1*40))/(5*4*3*2*1)$

[1] 0.1562935

dpois(5,4)

[1] 0.1562935
Can anybody help me review my result?
 A: I think you are not quite correct. Given that inter-arrival times are distributed $\mathrm{Exp(0.1)}$ and (presumably) independent, the number of arrivals in $40$ minutes follows a Poisson distribution with mean $4$, which seems to match what you have.
You now need the probability that at least five people arrive within this time. However, you appear to have worked out the probability that exactly $5$ arrive within this time.
A: Let $\{N(t):t\geqslant 0\}$ be a Poisson process with rate $\lambda>0$. Let $m$ be a positive integer and $T>0$ a fixed time. Then the probability that the first $m$ customers arrive during the first $T$ minutes is
\begin{align}
\mathbb P(N(T)\geqslant m) &= 1 - \sum_{k=0}^{m-1} \mathbb P(N(T)= k)\\
&= 1 - \sum_{k=0}^{m-1} e^{-\lambda T}\frac{(\lambda T)^k}{k!}\\
&= 1-\frac{\Gamma (m,T \lambda )}{\Gamma (m)}\\
&= 1 - \frac{\int_T^\infty t^{m-1}e^{-t}\ \mathsf dt}{(m-1)!}.
\end{align}
Substituting in $\lambda = \frac1{1/10}$, $m=5$ and $T=40$, this reduces to
$$
1-\frac{3232241203}3e^{-400} \approx 1.
$$
