Question

There are 27 presentations in total for a Conference. The probability of the Employee waking up to attend each presentation is .25.

(a) What is the distribution of the total number of presentations the Employee will attend?

(b) What is the probability the Employee will attend every presentation?

(c) What is the expectation and variance of the total number of presentations the Employee will attend during the entire conference?

(d) Assume the probability that it will rain is .3 and given that it rains, the probability of the Employee attending presentation drops to .05 and if it does not rain, the probability of him attending remains .25. What is the probability that it rained, given the Employee attended presentation?

Answer 1

Let X be the number of presentations the employee will attend out of the 27 presentations

The probability of the Employee waking up to attend each presentation is 0.25

a) Here the probability of attending each presentation is independent of the other probability.

Therefore X follows binomial distribution with parameters n = 27 and p = 0.25

where x = 0,1,2,...27

b)

Therefore the probability that the employee will attend every presentation is near to 0

c) The expectation of a binomial distribution = n.p

The variance of a binomial distribution = n.p.(1-p)

The expected number of presentations employee will attend = 27*0.25 = 6.75

The variance of the total number of presentations employee will attend = 27*0.25*0.75 = 5.0625

d) Let R be the event that it will rain and NR be the event that it will not rain

A be the event that employee will attend the presentation

Given P(R) = 0.3 ==> P(NR) = 1-0.3 = 0.7

P(A|R) = 0.05 and P(A|NR) = 0.25

P(A) = P(A|R).P(R) + P(A|NR).P(NR) Total Probability rule

P(A) = 0.05*0.3 + 0.25*0.7 = 0.19

  Bayes rule

Therefore probability that it rained given the employee attended the presentation = 0.0789