Question

A travel agent books passages on three different tours, with half of her customers choosing tour T1, one-third choosing tour T2, and the remaining one-sixth choosing tour T3. The agent has noted that 3/4 of those who take T1 return to book passage again, 2/3 of those who take T2 return, and 1/2 of those who take T3 return. If a customer does return, what is the probability that the person first went on tour T2

Answer 1

The probability that a returning customer first went on tour T2 is 8/19. This is found by dividing the number of returning customers from T2 by the total number of returning customers and simplifying.

Step-by-step explanation:

To find the probability that a returning customer went on tour T2, we need to apply the concept of conditional probability using Bayes' theorem. First, we calculate the fraction of returning customers for each tour. Then we find the probability that a returning customer came from T2.

Let's denote the total number of customers as 'C'. For tour T1, ½ C choose this tour, and ¾ of these return, so the returning customers from T1 are (½ C) * (¾) = 3/8 C. Similarly, for tour T2, (⅓ C) * (⅓) = 2/9 C return, and for T3, (⅙ C) * (½) = 1/12 C return.

The total number of returning customers is: 3/8 C + 2/9 C + 1/12 C = (9/24 + 8/24 + 2/24)C = 19/24 C.

The probability that a returning customer came from T2 is the number of T2 returnees over the total number of returnees: (2/9 C) / (19/24 C). The C's cancel out, simplifying to: (2/9) / (19/24) = (2/9) * (24/19) = 8/19.

Therefore, the probability that a returning customer first went on tour T2 is 8/19.