Editing Bayesian inference (section)

===Probability of a hypothesis===
{| class="wikitable floatright" style="font-size:100%;"
|+ [[Contingency table]]
! {{diagonal split header|<br />Cookie|Bowl}}
! #1<br />''H''<sub>1</sub> !! #2<br />''H''<sub>2</sub> !! rowspan="4" style="padding:0;"| !! <br />Total
|-
! Plain, ''E''
| '''30''' || 20 || '''50'''
|-
! Choc, ¬''E''
| 10 || 20 || 30
|-
! Total 
| 40 || 40 || 80
|-
| colspan="5"|''P''(''H''<sub>1</sub>|''E'') = 30 / 50 = 0.6
|}
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?

Intuitively, it seems clear that the answer should be more than a half, since there are more plain cookies in bowl #1. The precise answer is given by Bayes' theorem. Let <math>H_1</math> correspond to bowl #1, and <math>H_2</math> to bowl #2.
It is given that the bowls are identical from Fred's point of view, thus <math>P(H_1)=P(H_2)</math>, and the two must add up to 1, so both are equal to 0.5.
The event <math>E</math> is the observation of a plain cookie. From the contents of the bowls, we know that <math>P(E \mid H_1) = 30/40 = 0.75</math> and <math>P(E \mid H_2) = 20/40 = 0.5.</math> Bayes' formula then yields
<math display="block">\begin{align}
P(H_1 \mid E) &= \frac{P(E \mid H_1)\,P(H_1)}{P(E \mid H_1)\,P(H_1)\;+\;P(E \mid H_2)\,P(H_2)} \\
 \\
 \ & = \frac{0.75 \times 0.5}{0.75 \times 0.5 + 0.5 \times 0.5} \\
 \\
 \ & = 0.6
\end{align}</math>

Before we observed the cookie, the probability we assigned for Fred having chosen bowl #1 was the prior probability, <math>P(H_1)</math>, which was 0.5. After observing the cookie, we must revise the probability to <math>P(H_1 \mid E)</math>, which is 0.6.