**From:** Psy-Kosh (*psykosh@earthlink.net*)

**Date:** Sat Mar 29 2003 - 20:12:41 MST

**Next message:**John Stick: "Re: Bayesian story problem SPOILER WARNING: attempt at solution"**Previous message:**Eliezer S. Yudkowsky: "MATH: Bayesian story problem"**In reply to:**Eliezer S. Yudkowsky: "MATH: Bayesian story problem"**Next in thread:**John Stick: "Re: Bayesian story problem SPOILER WARNING: attempt at solution"**Reply:**John Stick: "Re: Bayesian story problem SPOILER WARNING: attempt at solution"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

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*> A couple of #SL4ers have asked me for this, so I'm posting it to
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*> SL4-the-mailing-list:
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*>
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*> Suppose you have a large barrel containing a number of plastic
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eggs. Some

*> eggs contain pearls, the rest contain nothing. Some eggs are
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painted

*> blue, the rest are painted red. Suppose that 40% of the eggs are
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painted

*> blue, 5/13 of the eggs containing pearls are painted blue, and 20%
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of the

*> eggs are both empty and painted red. What is the probability that
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an egg

*> painted blue contains a pearl?
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*>
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*> As a check on your calculations, the likelihood that a red egg is
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empty,

*> divided by the likelihood that an egg contains a pearl, equals
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*> approximately .51. To use this information in the problem, of
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course,

*> would be cheating.
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*>
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Spoiler space:

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Given:

P(B)=2/5

P(B|P)=5/13

P(~B,~P)=1/5

We have

1-P(B)=P(~B)=3/5

We can subtract the third item from the fourth to get the probability

that an egg contains a pearl and is red:

P(~B)-P(~B,~P)=P(~B,P)=2/5

Also we have the probability that an egg containing a pearl is red:

1-P(B|P)=P(~B|P)=8/13

Now, we are able to determine the total probability that an egg

contains a pearl:

P(~B|P)P(P)=P(~B,P) so P(P)=P(~B,P)/P(~B|P)=(2/5)/(8/13)

P(P)=13/20

So.... (Hail Bayes! All Hail Rationality! :)

P(P|B)=P(B|P)P(P)/P(B)

P(P|B)=(5/13)*(13/20)/(2/5)=1/10=10%

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