Total Probability Rule

Builds onExpected Value — if this page feels steep, start there.

P(B) = P(B\mid A)\,P(A) + P(B\mid A^{c})\,P(A^{c})

$P(B)$: the overall probability of event B, before knowing which scenario occurs
$P(B \mid A)$: the vertical bar reads 'given': the probability of B IF A happens
$A^{c}$: the little c means 'complement': the scenario where A does NOT happen
$P(B \mid A)\,P(A)$: the chance of arriving at B through the A route: B-if-A, weighted by how likely A is

Why it must be true

To find the overall probability of an event, slice the world into scenarios that cover everything and don't overlap — here, $A$ happens or it doesn't — and take a probability-weighted average of the event's chance in each slice.

This is the same mental move as expected value: analysts use it constantly ("probability the stock rises = chance it rises in a recession × chance of recession + chance it rises otherwise × chance of no recession"). It also supplies the denominator $P(B)$ in Bayes' formula.

The derivation

$A$ and its complement $A^c$ partition the sample space, so $B$ splits cleanly into two disjoint pieces:

P(B) = P(B \cap A) + P(B \cap A^{c})

Rewrite each joint probability with the multiplication rule $P(B \cap A) = P(B\mid A)P(A)$ :

P(B) = P(B\mid A)\,P(A) + P(B\mid A^{c})\,P(A^{c})

where $P(A^c) = 1 - P(A)$ .

When to reach for it

You know an event's probability inside each scenario and need its overall, unconditional probability.

Listen for

“if the economy expands … if it contracts …”“overall / unconditional probability”“probability under each scenario”

Back-of-the-envelope

Estimate it in your head first — then the calculator only confirms.

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The answer is a weighted blend, so it must land BETWEEN the two conditionals, closer to the likelier scenario's number. Anything outside that band is wrong.
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Use 100 imaginary cases: 60 expansions of which 80% survive (48) plus 40 recessions of which 35% survive (14) → 62 of 100 → 62%.

Traps in applying it

✗Weighting both branches by P(A) — the complement branch gets 1 − P(A).
✗Scenarios that overlap or don't cover every possibility.
✗Grabbing P(A|B) when the problem states P(B|A).

Limits & criticisms

The rule is exact, but only over a genuinely exhaustive, mutually exclusive partition — real scenario sets ('recession / no recession') compress a continuum into crude buckets. The output inherits whatever error lives in the scenario probabilities, which are usually the softest numbers in the room.

Where it came from

The rule was formalized in the classical era of probability, most completely by Pierre-Simon Laplace (Théorie analytique des probabilités, 1812), as the bookkeeping identity that makes conditional reasoning consistent: slice the world into scenarios, weight each, and nothing is double-counted or lost.

Today it is how analysts think in scenarios — "P(default) = P(default | recession)·P(recession) + …" — and it supplies the denominator of Bayes' formula, its constant companion.

Where it leads

Master this and the following come almost for free:

Bayes' Formula

Drill this formula Back to the library