# of people that go to a clinic follows a poisson distribution of 4 per day...... I just had an exam and I wanted to discuss a specific question on it. I will do my best to recall the question.
Suppose the number of people that go to a clinic follows a poisson distribution of 4 per day. What is the probability that no one will show up in the next 5 days? (Assume each day is independent). 
My answer was an extremely low probability.
 A: Already solved in the Comments, but for the permanent record on
this site--and the chance to illustrate a little more about
similar problems--here are two direct numerical answers.
1) Let $X \sim Pois(\lambda = 4).$ Then $P(X = 0) 
= e^{-\lambda}\lambda^0/0! = e^{-4} =  0.01831564$ is the probability of no patients showing on any one day. Because days are independent,
the probability of no patients on 5 successive days is (as per @Michael) $[P(X=0)]^5 = 2.0612 \times 10^{-9}.$
2) Let $Y$ be the number of patients in 5 days. 
Then $Y \sim Pois(\lambda_Y = 20).$ So the probability of
no patients in 5 days is $P(Y = 0) = e^{-20} = 2.0612 \times 10^{-9}.$
In working a Poisson problem, the first thing to do is to make
sure the rate is adjusted (if necessary) to match the domain
in which the random events occur. In (2) the adjusted rate for five days
is $\lambda_Y = 20$.
It is worthwhile noting that the two methods are about
equally easy for the question of getting no patients in five
days. But if the question were something like the probability
of getting three or fewer patients in five days, then method (1)
becomes a little awkward and method (2) remains fairly simple:
$$P(Y \le 3) = e^{-20}(1 + 20 + 20^2/2 + 20^3/6) = 3.2037 \times 10^{-6}$$
or in R software
 ppois(3, 20)
 ## 3.20372e-06

Still 'a very low' probability. However, $P(Y \le 25) = 0.8878$
is a reasonable question to ask in practice, with a larger answer, and not difficult to compute with software.
