Revise Conditional probability and Bayes' theorem for IB Mathematics: Analysis and Approaches (4.7)

The notation $P(A\mid B)$, read 'the probability of $A$ given $B$', asks a specific question: among the outcomes in which $B$ is already known to have happened, what fraction also satisfy $A$? Conditioning on $B$ throws away every outcome outside $B$ and treats $B$ as the new, smaller universe of possibilities. To measure $A$ inside that universe, we take the outcomes in both $A$ and $B$, namely $A\cap B$, and divide by the size of the new universe, $P(B)$.

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0.$

Independence has an intuitive meaning — learning that $B$ happened tells you nothing about whether $A$ happened — and an exact mathematical test. The two agree: if $B$ carries no information about $A$, then conditioning on $B$ should not move the probability, so $P(A\mid B)=P(A)$. Substituting this into the multiplication rule gives the symmetric form below, which is usually the cleanest to check because it needs no division.

$A\text{ and }B\text{ independent}\iff P(A\cap B)=P(A)\,P(B)\iff P(A\mid B)=P(A).$

Many questions describe events happening in stages, and give you conditional probabilities directly. A tree diagram is then the natural tool. The first set of branches shows the unconditional probabilities of the first stage; the second set shows conditional probabilities of the second stage given the first. Multiply along a path to get the joint probability of that whole path, and remember that the branches leaving any single node must sum to 1.

When an event $B$ can be reached by several mutually exclusive routes — say through $A$ or through its complement $A'$ — its total probability is the sum of the probabilities of all routes that end in $B$. This is the law of total probability, and it is exactly how we assemble the denominator we will need for Bayes' theorem.

$P(B)=P(B\mid A)\,P(A)+P(B\mid A')\,P(A').$ More generally, for a partition $A_1,A_2,\dots,A_n$ of the sample space, $P(B)=\sum_{i=1}^{n}P(B\mid A_i)\,P(A_i).$

Very often we know a probability in one direction but need it in the other. A test tells us $P(\text{positive}\mid\text{disease})$, but a patient wants $P(\text{disease}\mid\text{positive})$. Bayes' theorem performs this reversal. It follows in one line from the fact that the joint probability $P(A\cap B)$ can be written two ways, $P(A\mid B)P(B)=P(B\mid A)P(A)$; dividing by $P(B)$ isolates the conditional we want.

$P(A\mid B)=\frac{P(B\mid A)\,P(A)}{P(B)}=\frac{P(B\mid A)\,P(A)}{P(B\mid A)\,P(A)+P(B\mid A')\,P(A')}.$

Nothing changes when the evidence can arise from three or more mutually exclusive causes; the denominator simply gains more terms. If the causes partition the sample space as $A_1,\dots,A_n$, then for any particular cause $A_k$, $P(A_k\mid B)=\dfrac{P(B\mid A_k)P(A_k)}{\sum_{i} P(B\mid A_i)P(A_i)}$. The recipe is unchanged: the numerator is the one path through $A_k$, the denominator is the sum of all paths to $B$.