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 value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

step-by-step explanation: