Total Probability with Extra Conditioning¶

Theorem¶

Let $C$ be an event with $\mathbb{P}(C)>0$ and $\{B_i\}$ a finite or countable partition, and write $\mathbb{P}(A\mid B_i,C)$ for $\mathbb{P}(A\mid B_i\cap C)$ $C)$ . Then, omitting the terms with $\mathbb{P}(B_i\cap C)=0$ $0$ ,

\begin{aligned} \mathbb{P}(A\mid C)=\sum_i\mathbb{P}(A\mid B_i,C)\,\mathbb{P}(B_i\mid C). \end{aligned}

Analogous versions hold for expectations, PMFs, and PDFs.

Apply the Law of Total Probability within the probability law conditional on $C$ . Its case weights are $\mathbb{P}(B_i\mid C)$ rather than the unconditional weights.

Interpretation¶

Case-by-case probability calculations remain valid after some background information $C$ is fixed. Conditioning on $C$ gives a new probability function $\mathbb{P}(\,\cdot\mid C)$ $C)$ , and the ordinary law of total probability holds for it as for any other.

A probability tree inside a box labelled given C, a rainy day: the route branches into B1 and B2 with conditional probabilities 0.7 and 0.3, each followed by delay or on time; the delay branches combine to 0.29.

The tree in the figure lies entirely inside the box "given $C$ ". The first branches carry the case weights $\mathbb{P}(B_1\mid C)=0.7$ and $\mathbb{P}(B_2\mid C)=0.3$ $0.3$ , the second branches the delay probabilities $\mathbb{P}(A\mid B_1,C)=0.2$ and $\mathbb{P}(A\mid B_2,C)=0.5$ $0.5$ . Multiplying along the two delay paths gives $0.14$ and $0.15$ , and adding them gives $\mathbb{P}(A\mid C)=0.29$ $0.29$ , as in the first Example.

Both factors in each term must carry $C$ . The weights $\mathbb{P}(B_i\mid C)$ can differ from the unconditional $\mathbb{P}(B_i)$ : on rainy days a traveller may choose route $B_1$ more often than usual. Likewise $\mathbb{P}(A\mid B_i,C)$ is the delay probability on that route on a rainy day, not on an average day. Mixing the two, as in $\displaystyle \sum_i\mathbb{P}(A\mid B_i,C)\,\mathbb{P}(B_i)$ $B_{i}, C) P (B_{i})$ , is the typical slip; it is a valid rule only when every $B_i$ is independent of $C$ , so that $\mathbb{P}(B_i\mid C)=\mathbb{P}(B_i)$ $P (B_{i})$ .

The same pattern extends to the other conditioning rules: Bayes' rule with $C$ added to every term reads $\mathbb{P}(B_i\mid A,C)=\mathbb{P}(A\mid B_i,C)\,\mathbb{P}(B_i\mid C)/\mathbb{P}(A\mid C)$ $C)$ , and the version for expectations is Adam's law with extra conditioning.

Key Results¶

$\mathbb{P}(B_i\mid C)=\mathbb{P}(B_i\cap C)/\mathbb{P}(C)$ $C) / P (C)$ .
Only cases with $\mathbb{P}(B_i\cap C)>0$ need conditional probabilities $\mathbb{P}(A\mid B_i\cap C)$ $C)$ .

Examples¶

Within known rainy days $C$ , suppose $\mathbb{P}(B_1\mid C)=0.7$ $0.7$ , with delay probabilities $0.2$ and $0.5$ in the two routes. Then $\mathbb{P}(A\mid C)=0.7(0.2)+0.3(0.5)=0.29$ $0.29$ .
For a fair die conditioned to be even, partition into $B_1=\{X\le3\}$ and $B_2=\{X>3\}$ $3}$ . Then $\mathbb{P}(X=6\mid C)=0(1/3)+(1/2)(2/3)=1/3$ $1/3$ .

Prerequisites¶

Adam’s Law with Extra Conditioning

Sources¶

Nicky van Foreest · Probability Distributions · Week 6

Aliases: LOTP with extra conditioning, conditional law of total probability Symbols: intersection, sum

Last updated: 20 October 2026 03:02