Question

Please answer the following questions:

1) You are working at a call center at NYU calling alumni and trying to get them to donate money to the scholarship program (because God knows you the alumni didn’t drop enough of their money there already). Anyway, you realize that over the last month you average 9 donations per 4 hour shift. You are scheduled to work 7 hours Tuesday, what is the probability that get at least 16 donations?

2) You have taught high school for 22 years and have come to expect 3 ridiculous questions every 4 days. What is the probability that you get asked 16 ridiculous questions in the next 18 days

3) Suppose you dive into an Olympic sized swimming pool filled with peanut m&m’s and get 14 of them lodged in your clothes, ears nose, etc.
  a) What is the probability that you got 3 red, 3 green, 2 blue, 4 brown, 2 orange assuming that the colors are equally represented in the pool?
b) What is the probability that of the 14 marbles lodged in your pants and nose, that less than 5 of them are blue?

Answer 1

1. Expected no.of donations in the last month per 4 hour shift = 9

Expectednumber of donations per hour = 9/4 =2.25, Now if you are sheduled to work 7 hours on tuesday, Expected no. of donations you will get =7x2.25 =15.75

Let us suppose that no. of donations you will get follows poisson distribution ( since no one knows exactly how many donations we will get i.e indefinite) with Expected no.of donations m=the parameter of poisson distribution= 15.75

Now the required probability of getting atleast 16 donations in 7 hours duty on tuesday = P(X 16) = 1- P(X<16) = 1- 0.49173= 0.50873 ( from poisson probabilities)

2.Expected number of rediculous questions per day = 3/4 =0.75

Expected number of rediculous questions in 18 days = parameter of poisson distribution =0.75x 18 =13.5

now by poisson distribution probability of getting 16 rediculous quetions in 18 days = P(X = 16) =0.08331 by poisson probability function

3.if the colours are equally likely probability of getting any colour peanut m&m = 1/5 ( since there are 5 colours)

No.of peanut m&m lodged = 14

a) probability of getting 3 red= 0.25013 ( by binomial probability function)

  probability of getting 3 green = 0.25013

probability of getting 2 blue = 0.25013

  probability of getting 4 brown= 0.1719

probability of getting 2 orange = 0.25013

the required probability of getting 3 red , 3 green, 2 blue , 4 brown and 2 orange = product of the above probabilities= 0.0006728

b)probability of getting lessthan 5 of them are blue=P(X<5) = 0.1979 by cumulative probability of binomial distribution