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gulaghasi [49]
3 years ago
13

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 H

artford. 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.)
Mathematics
1 answer:
Law Incorporation [45]3 years ago
7 0

Answer:

The <em>probability </em>that it rains in Spain when hurricanes happen in Hartford is <em>0.1127</em>

Step-by-step 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

<em>P(R) = 1/10, </em>

<em>P(H | R) = 0.08, </em>

<em>P(H | R') = 0.07</em> and we want <em>P(R | H)</em>.

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

<em> where</em>

<em>P(H) = P(H | R)×P(R) + P(H | R')×P(R') </em>

<em />

Therefore,

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

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

<em>P(R | H) = 8 / 71</em>

<em>P(R | H) = 0.1127</em>

<em></em>

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

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