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Can someone help me with the following scenario, found on the Wikipedia page on Bayesian Inference:

Suppose there are two full bowls of cookies. Bowl #1 has 10 chocolate chip and 30 plain cookies, while bowl #2 has 20 of each. Our friend Fred picks a bowl at random, and then picks a cookie at random. We may assume there is no reason to believe Fred treats one bowl differently from another, likewise for the cookies. The cookie turns out to be a plain one. How probable is it that Fred picked it out of bowl #1?

I understand how to work out the probability when Fred picks one bowl and one cookie at random, which is:

\begin{align} P(H_1|E) &= \frac{P(E|H_1)\,P(H_1)}{P(E|H_1)\,P(H_1)\;+\;P(E|H_2)\,P(H_2)} \end{align}

\begin{align} = \frac{0.75 \times 0.5}{0.75 \times 0.5 + 0.5 \times 0.5} \ \ \ & = 0.6 \end{align}

What I would like to understand is how to work out the probability if Fred picks up a second cookie from the same bowl.

Can anyone help me with the formula?

Thanks

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You now want $P(E_1E_2|H_1)$ which is $\frac{30}{40}\times\frac{29}{39}$ and $P(E_1E_2|H_2)$ which is $\frac{20}{40}\times\frac{19}{39}$ etc. –  Dilip Sarwate Jan 4 '12 at 16:51

1 Answer 1

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The only thing that changes when Fred picks two cookies is that E is "Fred picks a plain cookie and then (say) another plain cookie". So we have $$ P(E|H_1) = \frac{30}{40} * \frac{29}{39} = \frac{29}{52} $$ $$ P(E|H_2) = \frac{20}{40} * \frac{19}{39} = \frac{19}{78} $$ These are simply the probability for each bowl of picking a plain cookie, and then picking another plain cookie from the remaining cookies. Now we can use the same formula, $$ P(H_1|E) = \frac{P(E|H_1)P(H_1)}{P(E|H_1)P(H_1) + P(E|H_2)P(H_2)} $$ $$ = \frac{\frac{29}{52} \times 0.5}{\frac{29}{52} \times 0.5 + \frac{19}{78} \times 0.5} = \frac{87}{125} $$ As a sanity check, we expect this to be greater than 0.6 since he picked another plain cookie and the first bowl has more plain cookies - and indeed, $87/125 = 0.696 > 0.6$ .

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