Customer arrivals at a bank are random and independent; the probability of an arrival in any one-minute period is the same as the probability of an arrival in any other one-minute period. A. What is the probability of exactly three arrivals in a one-minute period?
B. What is the probability of at least three arrivals in a one-minute period?

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

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation:

i think the amount would be $140.62