Conditional & Bayes formulas

Defined when $P(B)>0$. For fixed $B$, $P(\cdot \mid B)$ is itself a probability measure, so $P(A^c\mid B)=1-P(A\mid B)$.

The first form needs $P(B)>0$, the second needs $P(A)>0$. No independence is needed.

Used for sequential experiments such as draws without replacement. Needs $P(A_1\cap \cdots \cap A_{n-1})>0$, which makes every conditioning event positive.

Always valid. The second identity splits $A$ by whether $B$ occurs; it is the two-part case of the law of total probability below.

Always valid. Given any three of $P(A)$, $P(B)$, $P(A\cup B)$ and $P(A\cap B)$, it gives the fourth.

Always valid. Add the singles, subtract the pairs, add back the triple.

Sum over all values of the other variable. All cells of the table sum to 1.

Defined when $P(Y=y)>0$. Fix the column of the conditioning value, then divide each cell in it by that column's total. For each fixed $y$ these conditionals sum to 1 over $x$.

Each cell must equal the product of its row total and column total. One cell that fails is enough to show $X$ and $Y$ are dependent.

For jointly continuous $X,Y$ with joint density $f_{X,Y}$. Sums become integrals. The conditional density is defined where $f_Y(y)>0$. Densities are not probabilities: probabilities come from integrating them, and $P(X=x)=0$ for a continuous $X$.

The $B_i$ must be mutually exclusive and exhaustive with $P(B_i)>0$. The simplest partition is $\{B,B^c\}$: $P(A)=P(A\mid B)P(B)+P(A\mid B^c)P(B^c)$.

Follows from the multiplication rule written both ways. Needs $P(A)>0$ and $P(B)>0$.

$B_1,\dots ,B_n$ is a partition with positive probabilities, and $P(A)>0$. The denominator is the law of total probability, so the posteriors sum to 1 over $j$.

Posterior odds = likelihood ratio × prior odds. It needs $P(A\mid B^c)>0$, $0<P(B)<1$ and $P(A)>0$. Convert back with probability = odds / (1 + odds).

Valid when $E_1$ and $E_2$ are conditionally independent given $D$ and also given $D^c$. In the example above, two positive tests give $\frac{0.81\times 0.02}{0.81\times 0.02+0.0025\times 0.98}=\frac{0.0162}{0.01865}=\frac{324}{373}\approx 0.869$. Equivalently, apply Bayes once and use the first posterior as the prior for the second test.

The product form is the definition and needs no positivity condition. If $A$ and $B$ are independent, so are $A$ and $B^c$, $A^c$ and $B$, and $A^c$ and $B^c$.

$A$, $B$, $C$ are mutually independent when this holds AND every pair is independent: $P(A\cap B)=P(A)P(B)$, $P(A\cap C)=P(A)P(C)$, $P(B\cap C)=P(B)P(C)$. For $n$ events, the product rule must hold for every subcollection of two or more.

Events $A$ and $B$ are conditionally independent given $C$ when this holds, with $P(C)>0$. In general it neither implies nor is implied by ordinary independence.