Question

I need to verify my answers are correct.

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. Historical data indicates that 55% of all potential purchasers select a day visit, 25% choose a one-night visit, and 20% opt for a two-night visit. In addition, 10% of day visitors ultimately make a purchase, 25% of night visitors make a purchase, and 20% of those visiting for two nights make a purchase. Suppose a visitor is randomly selected

(a) What is the probability that the visitor makes a purchase?

(my answer was 0.1 * 0.55 + 0.25 * 0.25 + 0.2 * 0.2 = 0.1575)

(b) What is the probability that the visitor visited for two nights given that they made a purchase.

(my answer was found using bayes theorem: (0.25 * 0.2)/0.1575 = 0.31746

(c) What is the probability that the visitor visited for one night given that they did not make a purchase?

(my answer was: (1 - 0.25)(0.25)/(1 - 0.1575) = 0.22255

Thanks for your help

Answer 1

(a)

P(Makes purchase) = [P(Makes purchase/ Day visit) X P(Day visit)] + [P(Makes purchase/ Night visit) X P(Night visit)] + [P(Makes purchase/ Two Night visit) X P(Two Night visit)]

= (0.10 X 0.55) + (0.25 X 0.25) + (0.20 X 0.20)

= 0.055 + 0.0625 + 0.04 = 0.1575

So,

Answer is:

0.1575

(b)

P(Two nights/ Make purchase) = P(Two nights AND Make purchase)/ P(Made Purchase)

                                             = 0.2 X 0.2/0.1575

                                           = 0.2540

So,

Answer is:

0.2540

(c)

P(One night/ Not purchase) = P(One night AND Not purchase)/ P(Not purchase)

                                       = 0.25 X 0.75/(1 - 0.1575)

                                      = 0.1875/0.8425

                                      = 0.2226

So,

Answer is:

0.2226