Question

Suppose that it rains in Spain an average of once every 10 days, and when it does, hurricanes have a 8% chance of happening in Hartford. When it does not rain in Spain, hurricanes have a 7% chance of happening in Hartford. What is the probability that it rains in Spain when hurricanes happen in Hartford? (Round your answer to four decimal places.)

Answer 1

The probability that it rains in Spain when hurricanes happen in Hartford is 0.1127

Explanation:

This is a question where you use must use Bayes' Theorem.

The easiest way to do Bayes' type questions is to carefully define your terms.

Let R be the event that it is raining in Spain. R' is the event it isn't.

Let H be the event that it is hurricane in Hartford. H' is the event it isn't.

We know

P(R) = 1/10,

P(H | R) = 0.08,

P(H | R') = 0.07 and we want P(R | H).

Bayes Theorem says P(R | H) = [P(H | R)×P(R)] / P(H)

where

P(H) = P(H | R)×P(R) + P(H | R')×P(R')

Therefore,

P(R | H) = [P(H | R)×P(R)] / [P(H | R)×P(R) + P(H | R')×P(R')]

P(R | H) = [0.08 × 1/10] / [(0.08 × 1/10) + (0.07 × (1 - 1/10)]

P(R | H) = 8 / 71

P(R | H) = 0.1127

Therefore, the probability that it rains in Spain when hurricanes happen in Hartford is 0.1127.