The average rate of job submissions in a computer center is 2 per minute. 
Full Question: The average rate of job submissions in a computer
  center is 2 per minute. If it can be assumed that the number of
  submissions per minute has a Poisson distribution, calculate the
  probability that

(a) more than two jobs will arrive in a minute
$P(X>2) = 1- P(X≤ 2) = 1- [P(X=0)+P(X=1)+P(X=2)]$
and I got 0.3233235 as my answer. Is that correct?
Example: For $P(X=1),$ I did $((2^1)(e^-2))/1!$
(b) at least 30 sec will elapse between any two jobs
This is where I'm confused. A similar problem was answered using Exponential distribution in the book even though the question said this was a Poisson distribution problem.
I tried with Exponential and this is what I got:
Since 30 seconds is $1/2$ of 1 minute, $(1/2) * λ$ or $(1/2)*2 = 1$
$P(X ≥ 1/2) = 1 - P(x < 1/2) = 1 - (1- e^-1) = 0.3678794$
Please tell me if I used the correct distribution functions for both a and b, and if I answered them correctly. Thank you.
 A: Answers seem OK. Perhaps specifying the distributions more
clearly will help you have more confidence next time.
(a) Counting Poisson events: Let $X \sim Pois(\lambda = 2).$ Then $P(X > 2) =  0.3233236.$
From R statistical software:
1 - ppois(2, 2)   $ 'ppois' is Poisson CDF
## 0.3233236

(b) Exponential interarrival times: The waiting time $W$ for the next job is $W \sim Exp(rate = \lambda = 2).$
By the no-memory property, the starting time is not relevant, and $W$ is also
the wait between job arrivals (interarrival time). 
Then $P(W > .5) = 0.3678794.$ 
[Note that some books and software (including R)
parameterize the exponential distribution with the rate $\lambda,$ while others
use the mean $\mu = 1/\lambda.$ Please check the definition and the PDF and CDF
in your text.]
1 - pexp(.5, 2)  # 'pexp' is exponential CDF (rate parameter)
## 0.3678794

This also amounts to saying no Poisson event occurs in a time interval of 
length 1/2, for which the arrival rate is $\lambda^\prime = 1.$ Then
$P(X^\prime = 0) = 0.3678794.$
 dpois(0, 1)     # 'dpois' is Poisson PDF
 ## 0.3678794

