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

1)is not similar since corresponding sides are not in the same ratio

2)are similar since corresponding sides are in the same ratio and corresponding angles are equal

hope it will

#faazmessi

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i