Poisson: 3 purchases in an hour; based on 7 customers per hour with 65% probability of purchasing Disclaimer: It's a homework question worth a measly 1% of total grade, but I am actually interested in understanding the problem properly.

I've been given a question stating that 7 customers are expected to
  visit a website in any given hour, and that there is a 65% chance of a
  customer making a purchase. What is the probability that 3 purchases
  are made in an hour?

To my mind, probability of a purchase, therefore is: $\lambda = 7 \times 0.65 = 4.55$ 
Am I on the right track? Even if so, how to account for the notion of "3 purchases"?
I'm tempted to solve for: $1 - ([X=0] + [X=1] + [X=2]) = 0.831968$, but not sure that it answers the question posed.
Also, teacher said we may have invoke an identity: "$e$ to the power of $x$", but $e$ to the power of $3$ is $20.0855$, which leaves me rather baffled.
As I said... not so much interested in just getting the answer, but would like directions on how to solve the question.
Thanks and regards,
James

Yes, the process regarding the purchases will be a Poisson Process too with the new rate being $\lambda =  7 * 0.65 = 4.55$.
Let N(t) be the purchases made till time t. Using the formula for number of arrivals till time t, $Pr[N(t) = n] =  \frac{e^{-\lambda t}(\lambda t)^n}{n!}$ . In your case, $n$ is $3$ and $t$ is $1$. 
In case, you are wondering about how the formula for $Pr[N(t) = n]$ is derived, reading http://www.randomservices.org/random/poisson/ will be helpful.
