Question

1. Building traveler loyalty is a key strategic factor for a popular vacation rental platform. Several factors can deter repeat bookings, one of which is cancellation by the owner of the property. The data science team observes that 35% of travelers who have not had a reservation cancelled by an owner return to the site for repeat bookings. Travelers with a reservation that WAS cancelled by an owner repeat book at a 20% rate. If owners cancel 8% of the time, what is the overall repeat booking rate?

2. The data science team also seeks to understand the impact on customer loyalty with respect toanother type of cancellation: traveler cancellations. The repeat booking rate for owner cancellations is 20%;for traveler cancellations, 28%; and no cancellation, 36%. If the overall repeat booking rate is 34%, and theprobability of an owner cancellation is 8%, what is the probability of a traveler cancellation?

Answer 1

1. The overall repeat booking rate is 34.72%.

2. The probability of a traveler cancellation is 0.09 = 9%.

Explanation:

Bayes Theorem:

Two events, A and B.

`P(B|A) = (P(B)*P(A|B))/(P(A))`

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

Question 1:

36% of 100 - 8 = 92%(not canceled)

20% of 8%(canceled). So

`p = 0.36*0.92 + 0.2*0.08 = 0.3472`

The overall repeat booking rate is 34.72%.

Question 2:

We have that 34% is the sum of:

20% of 8%(repeat cancellation when the owner cancels).

28% of x(traveler cancelations).

36% of (100 - (8+x))%(no cancellation). So

`0.2*0.08 + 0.28x + 0.36(1 - (0.08+x)) = 0.34`

`0.016 + 0.28x + 0.36(0.92 - x) = 0.34`

`0.016 + 0.28x + 0.3312 - 0.36x = 0.34`

`0.08x = 0.3312 + 0.016 - 0.34`

`x = (0.3312 + 0.016 - 0.34)/(0.08)`

`x = 0.09`

The probability of a traveler cancellation is 0.09 = 9%.