# Bayesian inference

Assume two demographics $[F,M]$ and each person has a choice of attending only one of four different lectures $[A,B,C,D]$ all occurring at the same time so they can only attend one.

The following table is produced where all entries are mutually exclusive, each person belongs to only one of the eight cells.

\begin{array} {|r|r|r|} \hline & F & M \\ \hline A & 10 & 200 \\ \hline B & 20 & 10 \\ \hline C & 60 & 60 \\ \hline D & 900 & 10 \\ \hline \end{array}

The question now is:

• Given someone attended (only) lecture C, what is the probability they are F?

It may look like 50/50; but lectures people chose not to attend also gives information on their demographics. Lecture D is very popular with Females, so with all things being equal, not attending lecture D should increase the chances you are not Female. Similarly for Lecture A for Males.

Thinking along those lines, of trying to incorporate all information of which lectures they attended and did not attend; we used Bayesian Inference on Multiple Observations link on the formula

$P(F | \neg A , \neg B, C, \neg D) = 0.25 \\ P(M | \neg A , \neg B, C, \neg D) = 0.75$

It looks interesting that even though the demographics are evenly split for lecture C; the probabilities are actually (25%,75%) that you are either demographic given you attended lecture C. Is that the correct interpretation or procedure to solve the question? I feel that the 50/50 answer is missing information that can be used.

$P(F | \neg A , \neg B, C, \neg D) =\frac{ P( \neg A , \neg B, C, \neg D|F )\ \cdot \ P(F)}{P( \neg A , \neg B, C, \neg D )}$
$\hspace{4.4cm}=\frac{60/990\: \cdot\: 990/(990+280)}{120/(990+280)}$
$\hspace{4.4cm}=\frac{60/(990+280)}{120/(990+280)}$
$\hspace{4.4cm}=\frac{60}{120}$
$\hspace{4.4cm}=\frac{1}{2}$
The relevant information is the information you condition on - "people who attended lecture C". If you try to condition on $\neg A$ (etc) that adds nothing, since the lectures are mutually exclusive; as soon as you specify $C$ you already have that it's not-$A$. Bringing in the other lecture categories only scales numerator and denominator in a way that cancels out, leaving you again with the answer you could have got with less work.