CORE i5 7 Of those traveling to a regional statistics conference, 50% fly, 30% d
ID: 2922437 • Letter: C
Question
CORE i5 7 Of those traveling to a regional statistics conference, 50% fly, 30% drive, and 20% take a train or bus. Of those who fly, 90% have their travel expenses paid by their workplace whereas only 40% of those who drive and 30% of those taking the tain or bus have their travel expenses paid by their workplace. Suppose we randomly select one person arriving at the statistics conference. What is the probability that the person... a. (4 points) Drove to the conference and has their travel expenses covered? b. (4 points) Has their travel expenses covered? c. (4 points) Flew to the conference, given that their travel expenses were not covered?Explanation / Answer
Here we are given that:
P( fly ) = 0.5, P( drive ) = 0.3 and P( train ) =0.2
Also , we are given that P( paid | fly ) = 0.9, P( paid | drive ) = 0.4 and P( paid | train ) = 0.3
a) Probability that a person drove to the conference and has their travel expenses paid is computed as:
= P( drive ) P(paid | drive ) = 0.3*0.4 = 0.12
Therefore 0.12 is the required probability here.
b) Using addition law of probability we get:
P( paid ) = P( fly ) P(paid | fly ) + P( drive ) P(paid | drive ) + P( train ) P(paid | train )
P( paid ) = 0.9*0.5 + 0.3*0.4 + 0.3*0.2 = 0.63
Therefore 0.63 is the required probability here.
c) Now as we know from previous part that P( paid ) = 0.63. Therefore P( not paid ) = 1- 0.63 = 0.37
Given that the expenses were not paid, probability that the person flew to the conference is computed as: ( USing Bayes theorem )
P( fly | not paid ) = P( fly ) P( not paid | fly ) / P( not paid )
P( fly | not paid ) = 0.5*(1-0.9) / 0.37 = 0.1351
Therefore 0.1351 is the required probability here.