The number of people arriving for treatment at an emergency room can be modeled by a poisson process

with a rate parameter of five per hour.
(a) What is the probability that at exactly four arrivals occur during a particular hour.
(b) What is the probability that at least four people arrive during a particular hour?
(c) How many people do you expect to arrive during a 45-min period?

λ=5

a)
P(x=4) = e^-5 5^4 / 4! = 0.175467

b)
P( x >= 4) = 1-P(x=0)-P(x=1)-P(x=2)-P(x=3)

P(x=0) = e^-5 = 0.006738
P(x=1) = e^-5 5^1 /1! = 0.0336890
P(x=2) = e^-5 5^2 /2! = 0.084224
P(x=3) = e^-5 5^3 /3! = 0.1403739
P( x >=4) = 1- 0.006738 - 0.033689 - 0.084224-0.033689 = 0.8417

c)
5 per 60 minutes
In 60 minutes, 5 persons
in 1 minute 5/60 = 1/12 persons
in 45 minutes 45/12 = 3.75 persons

well,

(a)
the parameter of the Poisson process is a = 4 /hr
P( N = 4) = e^(-4) . 4^4/4!

(b) What is the probability that at least four people arrive during a particular hour?
P( N <= 4) = P(N=0) + P(N=1) + P(N=2) + P(N=3) + P(N=4)
with P(N=k) = e^4 4^k / k!

(c) How many people do you expect to arrive during a 45-min period?
the expectation of the time between two arrivals is :
1/4 = 15 min
therefore
the expected number of persons likely to come
in a 25 min time period is :
25/15 = 5/3

hope it' ll help !!

Hint:
X~Po(5) ----> P(X=k) = 5^k e^(-5)/k!