If number of customers are known, find probability wait time exceeds certain number The question is:

There is one checkout line and the average service time is 4 minutes per customer. There are 3 people in the queue ahead of you. What is the probability that your wait time will exceed 6 minutes?

I thought that the distribution was gamma with n = 3 and $\lambda$ = 1/4 but apparently I'm not correct. That is, I thought it was
$$\int^{\infty}_{6} {\frac{(1/4) e^{-x/4}(x/4)^{3-1}}{(3-1)!} dx} \approx 0.808847.$$
I'm thinking that there is a simpler way of computing this without (at least directly) using calculus because it is supposed to be an algebra-based statistics course.
EDIT. The reason why I think I am wrong is because the answer was marked incorrect for this online homework that I'm trying to help out a friend with. I wonder if Central Limit Theorem needs to be used here.
 A: I think you are right, if you are talking about the waiting time
to start being served. In R statistical software, where pgamma is the
CDF of $\mathsf{Gamma}(shape=3, rate=1/4):$
1 - pgamma(6, 3, 1/4)
## 0.8088468

But if you are talking about the waiting time before you are finished
being served, it would be:
1 - pgamma(6, 4, 1/4)
## 0.9343575

If that isn't the correct interpretation of the question, maybe you can share the 'answerbook' answer, or say
why you think you answer is wrong.

Note: I can think of another possibility: Even in these days of cheap computation, some elementary probability books persist in
using barely appropriate normal approximations in such problems, in
order to avoid dealing with gamma PDFs and CDFs.
The waiting time until you start being served has $\mu = 12$ and $\sigma = \sqrt{48}.$ The waiting time until you finish has $\mu = 16$ and $\sigma = 8.$
Then "normal approximations" are:
1 - pnorm(6, 12, sqrt(48))
## 0.8067619
1 - pnorm(6, 16, 8)
## 0.8943502

The figure shows the density functions of $\mathsf{Gamma}(4, 1/4)$
and $\mathsf{Norm}(16,8)$. Any resemblance between the correct gamma
probability and the approximate normal probability seems mostly accidental.

A comparison the density functions of $\mathsf{Gamma}(3, 1/4)$
and $\mathsf{Norm}(12,\sqrt{48})$ happens to work a little better. 

