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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?

Sagot :

Answer:

1. The overall repeat booking rate is 34.72%.

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

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]

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

[tex]p = 0.36*0.92 + 0.2*0.08 = 0.3472[/tex]

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

[tex]0.2*0.08 + 0.28x + 0.36(1 - (0.08+x)) = 0.34[/tex]

[tex]0.016 + 0.28x + 0.36(0.92 - x) = 0.34[/tex]

[tex]0.016 + 0.28x + 0.3312 - 0.36x = 0.34[/tex]

[tex]0.08x = 0.3312 + 0.016 - 0.34[/tex]

[tex]x = \frac{0.3312 + 0.016 - 0.34}{0.08}[/tex]

[tex]x = 0.09[/tex]

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

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