We are given that an event is expected to occur 20 times a day We are given that an event is expected to occur 20 times a day. 
I think that this event is a continuous random variable, and there should be really no effect of these events on one another.
How would I start to find the probability that this event occurs only 15 or fewer times on a given day and how would I find the probability between different times of these events occuring?
From what I've gathered so far there will be roughly 1.2 hours between events by using E(T) = 1/lambda.
I'd prefer not to be given answers and just the best way to start this problem out, I've been trying to find the formulas I would need through google but have only been able to find homework assignments from certain colleges.
Thank you!
 A: Here are indications of the arguments that lead to answers for both of you questions.
Probability of 15 or fewer in a day. If $X \sim \mathsf{Pois}(\lambda = 20)$ is the number of events per day,
then $$P(X \le 15) = \sum_{x=0}^{15} \frac{e^{-20}20^x}{x!} = 0.1565,$$
evaluated using R statistical software as:
 sum(dpois(0:15, 20))  # 'dpois' is Poisson PDF
 ## 0.1565131

 ppois(15, 20)         # 'ppois' is Poisson CDF
 ## 0.1565131

This may be roughly approximated by a normal distribution:
$$P(X \le 15) = P(X < 15.5) = 
P\left(\frac{X-\lambda}{\sqrt{\lambda}} < \frac{15.5-20}{\sqrt{20}}\right)
\approx P(Z < -1.006) = 0.1572,$$
where $Z$ is standard normal, and the probability can be found from 
printed tables.

Expected interarrival time. Starting at time $t = 0,$ the probability that the waiting time $W$ for
the first event exceeds $t$ is $P(W > t)$ is the same as the probability
 $P(X_t = 0)$ of no event in $(0, t),$ where 
$X_t \sim \mathsf{Pois}(\lambda_t = t\lambda).$
Then $P(W > t) = P(X_t = 0) = e^{-\lambda t},$ so that the CDF of
$W$ is $F_W(t) = P(W \le t) = 1 - e^{-\lambda t},$ for $t > 0.$
Then by differentiation, the density function of $W$ is
$f_W(t) = \lambda e^{-\lambda t},$ for $t > 0.$ This is the density
function of $W \sim \mathsf{Exp}(rate = \lambda).$ And $E(W) = 1/\lambda$
can be found using integration by parts in $E(W) = \int_0^\infty tf_W(t)\,dt.$
By the no-memory property of the exponential distribution the waiting
time for the first event starting at $t=0$ is the same as the waiting
time for the $(i+1)$st event starting at the time of the $i$th event.
