Conditional Probability

Knowing that $B$ has occurred rules out every outcome outside $B$. So instead of asking what fraction of the whole sample space is $A$, you ask what fraction of $\text{B}$ is also $A$.

That gives the formula directly: the outcomes in both, divided by the outcomes in $B$. Dividing by $P(B)$ is what rescales the reduced space so its probabilities still sum to 1.

On a Venn diagram, $P(A\mid B)$ means covering everything outside the $B$ circle and asking what proportion of what remains lies in the overlap.

A two-way table is often the fastest route. The conditioning event selects a row or column, and the answer is one cell divided by that row or column total — rather than by the grand total.

Getting the denominator right is the whole skill. Dividing by the grand total gives $P(A\cap B)$, not $P(A\mid B)$, and the two answers can look equally plausible.

The numerator $P(A\cap B)$ is the same for both directions; only the denominator changes. So $P(A\mid B)$ and $P(B\mid A)$ are equal only when $P(A)=P(B)$.

When the two events have very different overall probabilities, the conditionals can differ dramatically. If a rare condition is almost always accompanied by a common symptom, then $P(\text{symptom}\mid \text{condition})$ is near 1 while $P(\text{condition}\mid \text{symptom})$ may be tiny, because the symptom has many other causes.

The general relationship between the two directions is obtained by writing $P(A\cap B)$ two ways and equating.

When events happen in sequence, the multiplication law extends: $P(A\cap B)=P(A)\times P(B\mid A)$. Each stage is conditioned on everything before it.

Selection $\text{without replacement}$ is the standard example. Removing an item changes both the favourable count and the total for the next draw, so the second probability is genuinely conditional.

With replacement, the conditional probability equals the unconditional one, which is exactly the definition of independence in the next lesson.