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 April 5th, 2013, 10:59 PM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Check my answer please - conditional probability (bayes) hey all, could you give comments and check my answer? It doesn't show in the textbook. OriaG is randomly choosing a city to visit. city A, city B, and city C. The chance it will rain on city A is $\frac{1}{3}$ The chance it will rain on city B is $\frac{1}{4}$ The chance it will rain on city C is $\frac{1}{6}$ When OriaG arrived at his city of choice, it was raining. What is the chance OriaG chose to visit city C? My solution: event A - OriaG chose to visit city A. same for events B and C. event D - It's raining somewhere event D|A - it's raining in city A. same for events D|B and D|C. event C|D - OriaG chose city C and it's raining. $P(D)= \frac{P(D|A)+P(D|B)+P(D|C)}{3} = \frac{\frac{1}{3}+\frac{1}{4}+\frac{1}{6}}{3} = \frac{1}{4}$ we we're given $P(D|C)= \frac{1}{6}$ and we are asked to calculate $P(C|D)$, use Bayes formula: $P(C|D)= \frac{P(D|C)*P(C)}{P(D)} = \frac{\frac{1}{6}*\frac{1}{3}}{\frac{1}{4}} = \frac{2}{9}$ Final answer: The chance OriaG chose a rainy city and it was C is $\frac{2}{9}$
 April 6th, 2013, 11:27 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Check my answer please - conditional probability (bayes) Yes, that is what I get doing the problem a different way: Imagine a set of 36 "(city, weather)" pairs. 12 each refer to A, B, and C. Of the "A" pairs, 1/3, or 4, are (A, rain). Of the "B" pairs, 1/4, or 3, are (B,rain). Of the "C" pairs, 1/6, or 2, are (C, rain). That' makes a total of 9 "(city, weather)" pairs that have "rain". Of those, 2 are "City C" so the probability he/she went to city C given that it was raining is $\frac{2}{9}$.

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