Answer:
0.2240418 ; 0.57681
Step-by-step explanation:
Given the information above :
A) What is the probability of exactly three arrivals in a one-minute period?
Using poisson probability function :
p(x ; m) = [(m^x) * (e^-m)] / x!
Here, m = mean = 3, x = 3
P(3 ; 3) = [(3^3) * (e^-3)] / 3!
P(3;3) = [27 * 0.0497870] / 6
= 1.3442508 / 6
= 0.2240418
B) What is the probability of at least three arrivals in a one-minute period?
Atleast 3 arrivals
X >= 3 = 1 - [p(0) + p(1) + p(2)]
P(0 ; 3) = [(3^0) * (e^-3)] / 0! = (1 * 0.0497870) / 1 = 0.0497870
P(1 ; 3) = [(3^1) * (e^-3)] / 1! = (3 * 0.0497870) / 1 = 0.1493612
P(2 ; 3) = [(3^2) * (e^-3)] / 2! = (9 * 0.0497870) / 2 = 0.2240418
1 - [0.0497870 + 0.1493612 + 0.2240418]
1 - 0.42319 = 0.57681
