# Bayes Rule question

20 percent of blue eyed people have brown hair and thirty percent of non blue eyed people have brown hair. 10 percent of people are blue eyed. A random person has brown hair. What is the probability that person has blue eyes.

0.2 BE have BH and 0.3 BE^C have BH P(BE) = 0.1 P(BE^C) = 0.9 P(BH|BE) = 0.2 P(BH|BE^c) = 0.3 P(BH) = P(BH|BE) + P(BH|BE^c) = 0.2 + 0.3 = 0.5

So P(BE|BH) = P(BH|BE)P(BE)/(P(BH)) = (0.2)(0.1)/(0.5) = 0.4. Is this correct?

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You've asked quite a few questions on this site now. It would be useful for you (and helpful for others) to learn just enough $\LaTeX$ to format the math in your questions nicely. It is (much) easier on the eyes and so doing this will undoubtedly attract more respondents to your questions. Cheers. –  cardinal Sep 25 '11 at 18:38

## 2 Answers

In the denominator, how did you get $P(BH) = 0.5$? You have to calculate this given the other numbers $P(BH) = P(BH|BE)P(BE)+P(BH|BE^c)P(BE^C) = 0.2*0.1+0.3*0.9 = 0.29$

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In addition to previous answer: It looks like you misunderstand the law of total probability

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