Properties of conditional probability

Let $S$ be the sample space and $B$ an event with $P(B) \neq 0$. Three properties describe how $P(\cdot \mid B)$ behaves.

Whole space. Conditioning on $B$ turns $B$ itself into the "new" certain event, so both $S$ and $B$ get probability $1$ when conditioned on $B$:

Addition law under conditioning. For any two events $A$ and $C$ of $S$,

where: the subtracted term corrects for outcomes counted twice, exactly as in the ordinary addition rule. If $A$ and $C$ are mutually exclusive ($A\cap C = \varnothing$), the correction vanishes and $P\big((A\cup C)\mid B\big) = P(A\mid B) + P(C\mid B)$.

Complement. For any event $A$,

where: $A'$ is the complement of $A$ in $S$, and $B$ — the conditioning event — never changes; only $A$ gets complemented. These three properties together say precisely that $P(\cdot \mid B)$ satisfies the same axioms as an ordinary probability function, just measured relative to $B$ instead of $S$: it is non-negative, assigns $1$ to the certain event (now $S$ or $B$), and adds up correctly over unions. That is why every standard identity of probability — addition rule, complement rule, and beyond — carries over automatically the moment you replace "probability of an event" with "probability of an event given $B$." A useful consequence for numerical problems: for events $A_1, A_2$ conditioned on $B$,

with equality exactly when $A_1$ and $A_2$ do not overlap within $B$. The figure below shows why: inside the reduced sample space $B$, the pieces of $A_1$ and $A_2$ either sit apart (equality, mutually exclusive) or overlap (strict inequality, needing the correction term).

In a group of 100 students, 60 like Cricket ($C$), 45 like Football ($F$), and 30 like both. A student is picked at random and is known to like Cricket. Find $P(F' \mid C)$, the probability the student does not like Football, given the student likes Cricket.

Step 1. Set up the probabilities. $P(C) = \tfrac{60}{100} = 0.6$, $P(F) = \tfrac{45}{100} = 0.45$, $P(C\cap F) = \tfrac{30}{100} = 0.3$.

Step 2. Find $P(F\mid C)$ from the definition. $$P(F\mid C) = \frac{P(C\cap F)}{P(C)} = \frac{0.3}{0.6} = 0.5.$$

Step 3. Apply the complement property. $$P(F' \mid C) = 1 - P(F\mid C) = 1 - 0.5 = \mathbf{0.5}.$$ This matches direct counting: of the 60 Cricket-lovers, 30 also like Football, so exactly 30 — half — do not.

Competitive stretch

Two dice are thrown; let $B$ be "the sum is 8," $A_1$ be "the first die is even," and $A_2$ be "the first die shows 5." Sum $= 8$ occurs for $(2,6),(3,5),(4,4),(5,3),(6,2)$, so $P(B) = \tfrac{5}{36}$. Within $B$, $A_1$ gives $(2,6),(4,4),(6,2)$ so $P(A_1\mid B) = \tfrac{3}{5}$, and $A_2$ gives $(5,3)$ so $P(A_2\mid B) = \tfrac{1}{5}$. Since $A_1$ and $A_2$ cannot both hold (first die cannot be even and equal to 5), they are mutually exclusive, so the addition law gives $P\big((A_1\cup A_2)\mid B\big) = \tfrac{3}{5}+\tfrac{1}{5} = \mathbf{4/5}$ — which matches direct counting of the 4 outcomes $(2,6),(4,4),(6,2),(5,3)$ out of the 5 in $B$.

• Applying the addition law without subtracting the intersection term when the events are not mutually exclusive — the correction $-P((A\cap C)\mid B)$ is easy to forget.
• Misapplying the complement property to the wrong event: $P(A'\mid B) = 1 - P(A\mid B)$, not $1 - P(A\mid B')$ or $1 - P(A'\mid B')$ — only $A$ is complemented, $B$ stays fixed.
• Forgetting $P(B\mid B) = 1$ always, so $P(B'\mid B) = 0$ trivially — a degenerate case that still gets tested.
• Using these properties when $P(B) = 0$, where conditional probability is undefined.
• Treating the addition law as new content to memorize separately from the ordinary addition rule, instead of seeing it as the same identity divided through by $P(B)$ — remembering the reason it holds makes it far harder to misquote under exam pressure.

Ask the tutor: Want to see all three properties combined in one multi-part problem, or need help spotting which property a given question is testing? Ask the AI tutor for a guided walkthrough.