Conditional probability — P(A - B) = P(A∩B) - P(A)

WHAT is being defined

• $A\cap B$ = "both $A$ and $B$" (the overlap).
• We ==restrict the universe to $B$== and measure how much of $B$ is taken up by $A$.

WHY the formula must look like this (derivation from scratch)

Suppose outcomes are equally likely, with $N$ total outcomes.

• $P(B) = \dfrac{n(B)}{N}$ where $n(B)$ = number of outcomes in $B$.
• Once we know $B$ happened, only the $n(B)$ outcomes in $B$ are still possible. The new sample space is $B$.
• Among these, the ones that also give $A$ number $n(A\cap B)$.

So by the basic "favourable / total" rule inside the shrunken world: $P(A|B) = \frac{n(A\cap B)}{n(B)}$

Now divide top and bottom by $N$ (the trick that removes the "equally likely" assumption): $P(A|B) = \frac{n(A\cap B)/N}{n(B)/N} = \frac{P(A\cap B)}{P(B)}$

The multiplication rule (rearranged form)

Multiply both sides by $P(B)$: $P(A\cap B) = P(B)\,P(A|B) = P(A)\,P(B|A)$

Independence falls out for free

If knowing $B$ tells you nothing about $A$, then $P(A|B) = P(A)$. Substitute: $P(A\cap B) = P(B)\,P(A|B) = P(B)P(A)$ which is the independence condition. So: $A,B \text{ independent} \iff P(A|B)=P(A) \iff P(A\cap B)=P(A)P(B)$

Worked examples

Common mistakes (Steel-manned)

Flashcards

Connections

• Independence of events — special case where $P(A|B)=P(A)$.
• Multiplication rule of probability — the rearranged form.
• Bayes' Theorem — flips $P(A|B)$ into $P(B|A)$ correctly.
• Law of total probability — sums conditionals over a partition.
• Sample space and events — the "universe" we shrink.
• Tree diagrams — visual chaining of conditional steps.

Concept Map

Hinglish (regional understanding)