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1.4 Conditional Probability

The probability of an event depends not just on the experiment itself but on other information you are given about the experiment. Conditional probability forms a framework in which this additional information can be incorporated.

If $A$ and $B$ are two events then, as long as $\operatorname{\mathsf{P}}\left({B}\right)>0$ , the conditional probability of $A$ given $B$ is written as $\operatorname{\mathsf{P}}\left({A|B}\right)$ and calculated from

\displaystyle\operatorname{\mathsf{P}}\left({A|B}\right)=\operatorname{\mathsf% {P}}\left({A\cap B}\right)/\operatorname{\mathsf{P}}\left({B}\right).

Note that this is a probability since:

Positivity: $\operatorname{\mathsf{P}}\left({A\cap B}\right)\geq 0$ and $\operatorname{\mathsf{P}}\left({B}\right)>0$ so $\operatorname{\mathsf{P}}\left({A|B}\right)\geq 0$ .
Finiteness: $\operatorname{\mathsf{P}}\left({\Omega|B}\right)=\operatorname{\mathsf{P}}% \left({\Omega\cap B}\right)/\operatorname{\mathsf{P}}\left({B}\right)=1$ .
Additivity: Let $A_{1}$ and $A_{2}$ be disjoint sets $(A_{1}\cap A_{2}=\emptyset)$ . Then $A_{1}\cap B$ and $A_{2}\cap B$ are also disjoint, so (law of total probability)

$\displaystyle\operatorname{\mathsf{P}}\left({[A_{1}\cup A_{2}]\cap B}\right)={% \color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({[A_{1}\cap B]\cup[A_{2% }\cap B]}\right)=\operatorname{\mathsf{P}}\left({A_{1}\cap B}\right)+% \operatorname{\mathsf{P}}\left({A_{2}\cap B}\right)}.$

Dividing by $\operatorname{\mathsf{P}}\left({B}\right)$ gives: $\operatorname{\mathsf{P}}\left({A_{1}\cup A_{2}|B}\right)=\operatorname{% \mathsf{P}}\left({A_{1}|B}\right)+\operatorname{\mathsf{P}}\left({A_{2}|B}\right)$ .

When events $A$ and $B$ are independent $\operatorname{\mathsf{P}}\left({A|B}\right)=\operatorname{\mathsf{P}}\left({A}\right)$ .

It is often easiest to evaluate $\operatorname{\mathsf{P}}\left({A\cap B}\right)$ using $\operatorname{\mathsf{P}}\left({A\cap B}\right)=\operatorname{\mathsf{P}}\left% ({A|B}\right)\operatorname{\mathsf{P}}\left({B}\right)=\operatorname{\mathsf{P% }}\left({B|A}\right)\operatorname{\mathsf{P}}\left({A}\right)$ .

Bayes theorem inverts the ordering of conditioning for events $A$ and $B$ :

\displaystyle\operatorname{\mathsf{P}}\left({B|A}\right)=\operatorname{\mathsf% {P}}\left({A|B}\right)\operatorname{\mathsf{P}}\left({B}\right)/\operatorname{% \mathsf{P}}\left({A}\right).