"The probability of $A$, given that $B$ has already happened." The condition $B$ shrinks the sample space down to only the outcomes where $B$ is true — everything outside $B$ becomes irrelevant.

Example: from $\{1,\ldots,9\}$, $P(\text{multiple of }3)=\tfrac39=\tfrac13$. But given the number is even (new universe $\{2,4,6,8\}$), $P(\text{multiple of }3\mid\text{even})=\tfrac14$ — a different answer, because the denominator changed.

This is just the intuitive method written algebraically: $P(B)$ is the new (restricted) denominator, and $P(A\cap B)$ is how much of $A$ survives inside that restriction.

The general rule always holds: $P(A\cap B)=P(A)\times P(B|A)$. If the events are independent, knowing $A$ happened changes nothing, so $P(B|A)=P(B)$ — and the formula collapses to the familiar $P(A\cap B)=P(A)\times P(B)$. Independent events are just a special case of the general rule, not a separate one.

The second set of branches must be adjusted: the total decreases by 1, and whichever outcome was picked first also loses one from its own count.

Example: bag has 6 red, 4 green (10 total), no replacement.

• First pick: $P(R_1)=\tfrac6{10}$, $P(G_1)=\tfrac4{10}$.
• If red was picked first: 5 red, 4 green left (9 total) → $P(R_2|R_1)=\tfrac59$.
• If green was picked first: 6 red, 3 green left (9 total) → $P(G_2|G_1)=\tfrac39$.

Multiply along branches for "and"; add between branches for "or".

Don't reuse the forward branch probability directly — apply the formula properly, since $P(A|B)\ne P(B|A)$ in general.

Worked exam-style question: Same bag (6 red, 4 green, no replacement). Given the second marble was green, find the probability the first was red.

1. Target: $P(R_1|G_2)=\dfrac{P(R_1\cap G_2)}{P(G_2)}$.
2. Numerator: $P(R_1\cap G_2)=\tfrac6{10}\times\tfrac49=\tfrac{24}{90}$.
3. Denominator — $G_2$ can happen two ways: $P(R_1\cap G_2)+P(G_1\cap G_2)=\tfrac{24}{90}+\left(\tfrac4{10}\times\tfrac39\right)=\tfrac{24}{90}+\tfrac{12}{90}=\tfrac{36}{90}$.
4. $P(R_1|G_2)=\dfrac{24/90}{36/90}=\dfrac{24}{36}=\dfrac23$.

Same restriction idea: the "given" event's row/column/circle total becomes your denominator; the overlap with the other event becomes your numerator.

Two-way table example (55 students): $P(\text{Spanish}\mid\text{Year 13})$ uses only the Year 13 row (total 30), with Spanish-and-Year-13 (5) on top: $\tfrac{5}{30}=\tfrac16$. Note $P(\text{Year 13}\mid\text{Spanish})$ uses the Spanish column total (20) instead: $\tfrac{5}{20}=\tfrac14$ — a completely different number, confirming $P(A|B)\ne P(B|A)$.

Check either equivalent condition:

Worked example: $P(A)=0.5$, $P(B)=0.6$, $P(A\cup B)=0.8$. Find $P(A\cap B)$ via the addition rule: $0.8=0.5+0.6-P(A\cap B)\Rightarrow P(A\cap B)=0.3$. Compare: $P(A)\times P(B)=0.5\times0.6=0.3$. Since these match, $A$ and $B$ are independent.

• Confusing $P(A\cap B)$ (divide by the overall total) with $P(A|B)$ (divide by the restricted total) — "given" is an instruction to change your denominator.
• Swapping $P(A|B)$ for $P(B|A)$.
• Forgetting that $P(B)$ in the denominator of a reverse-conditional problem usually has to be built from two branches added together, not read straight off one.
• Assuming events are independent without actually checking the test.