A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. Historical data indicates that 20% of all potential purchasers select a day visit, 50% choose a one-night visit, and 30% opt for a two-night visit. In addition, 10% of day visitors ultimately make a purchase, 30% of one night visitors buy a unit, and 20% of those visiting for two nights decide to buy. Suppose a visitor is randomly selected and is found to have made a purchase. How likely is it that this person made a day visit? A one-night visit? A two-night visit?

Solution: Step1: Here based on the given data 20% of all potential purchasers select a day visit, 50% choose a one-night visit, and 30% opt for a two night visit In addition, 10% of day visitors ultimately make a purchase, 30% of one night visitors buy a unit, and 20% of those visiting for two nights decide to buy. A visitor is randomly selected and is found to have a made a purchase. Step2: a) We have to find the probability that selected person made a day visit. Here it is given that among 20% of day visitors 10% makes a purchase. Also among 50% of one- night visitors 30% purchase a unit. And among 30% of two- night visitors 20% makes a purchase. So based on this given information we can make probability table. Purchase yes No total Day visit .02 .18 .2 Visit .15 .35 .5 One-night visit .06 .24 .3 Two-night visit It is given that the randomly selected person made a purchase. Let A be the event that selected visitor made a purchase.And also lets define the events E1: The selected person made a day visit E2: The selected visitor made a one-night visit E3: The selected visitor made a two-night visit. So by bayes theorem the probability that selected the selected visitor made a day- visit. P(selected visitor made a day- visit given that visitor made a purchase)= (E1/A)= P(A/E1) × P(E1) P(A/E1)× P(E1)+P(A/E2)×P(E2)+P(A/E3)×P(E3) Here from the table P(A/E1)= .02 P(A/E2)= .15 P(A/E3) = .06 And also P(E1) = .2 P(E2) = .5 P(E3) = .3 So (E1/A) = .02×.2 .15×.5+.02×.2.+.06×.3 = 0.041 Step3 : b) P(selected visitor made a one-night visit given that visitor made a purchase)= P(A/E2) × P(E2) P(E2/A)= P(A/E1)× P(E1)+P(A/E2)×P(E2)+P(A/E3)×P(E3) So .15×.25 (E2/A) = .15×.5+.02×.2.+.06×.3 = 0.773 c) P(selected visitor made a two-night visit given that visitor made a purchase)= P(A/E3) × P(E3) P(E3/A)= P(A/E1)× P(E1)+P(A/E2)×P(E2)+P(A/E3)×P(E3) So (E3/A) = .06×.3 .15×.5+.02×.2.+.06×.3 = 0.1855