An airline offers discounted advancepurchase fares to custom

An airline offers discounted “advance-purchase” fares to customers who buy more than 30 days before travel and charges “regular” fares for tickets purchased during those last 30 days. Company records show that 60% of customers take advantage of the “advance-purchase” fares and thus 40% purchase “regular” fares. The “no-show” rate among customers who purchased a “regular” fare is 30%, but only 5% of customers with “advance-purchase” fares are “no-shows”.

A) What is the probability that a customer is a “no-show”?

B) Given that a customer is a “no-show”, what is the probability that they had an “advance-purchase” fare?

Solution

A)
P(customers who take advantage of the “advance-purchase”) = 0.6
P(customers who purchase “regular”) = 0.4

P(customers who take advantage of the “advance-purchase” and no-show) = 0.05 * 0.6 = 0.03
P(customers who purchase “regular” and \"no-show\") = 0.3*0.4 = 0.12

Therefore,
P(customer is a “no-show”) = 0.03+0.12 = 0.15 Answer

B)
P(customer is a “no-show”) = 0.15
P(customers who take advantage of the “advance-purchase” and no-show) = 0.03

Therefore,
P(customer had an “advance-purchase” / customer is a “no-show”)
= P(customers who take advantage of the “advance-purchase” and no-show)/P(customer is a “no-show”)
= 0.03 / 0.15
= 0.2 Answer