# Re: Bayesian story problem SPOILER WARNING: attempt at solution

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

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> A couple of #SL4ers have asked me for this, so I'm posting it to
> SL4-the-mailing-list:
>
> Suppose you have a large barrel containing a number of plastic
eggs. Some
> eggs contain pearls, the rest contain nothing. Some eggs are
painted
> blue, the rest are painted red. Suppose that 40% of the eggs are
painted
> blue, 5/13 of the eggs containing pearls are painted blue, and 20%
of the
> eggs are both empty and painted red. What is the probability that
an egg
> painted blue contains a pearl?
>
> As a check on your calculations, the likelihood that a red egg is
empty,
> divided by the likelihood that an egg contains a pearl, equals
> approximately .51. To use this information in the problem, of
course,
> would be cheating.
>

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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