Why do independent probabilities multiply?

For events $A$ and $B$ with $P(A)>0$, conditional probability is defined by $P(B\mid A)=\frac{P(A\cap B)}{P(A)}.$ Multiplying both sides by $P(A)$ gives the multiplication rule $P(A\cap B)=P(A)P(B\mid A).$ This identity does not require independence: the second factor is the probability of $B$ after learning that $A$ occurred. If $P(A)=0$, then $P(A\cap B)=0$ as well because the intersection is contained in $A$.

By definition, $A$ and $B$ are independent when $P(A\cap B)=P(A)P(B).$ If $P(A)>0$, this is equivalent to $P(B\mid A)=P(B).$ In plain language, knowing that $A$ occurred does not change the probability of $B$. Substituting this value into the general rule gives $P(A\cap B)=P(A)P(B).$ The probabilities multiply because each independent step keeps its probability after the preceding steps are known.

Let $H_1$ mean heads on the first flip and $H_2$ mean heads on the second. The flips are independent and each event has probability $1/2$, so $P(H_1\cap H_2)=\frac12\cdot\frac12=\frac14.$ We can check the result by listing the ordered outcomes: $HH,\ HT,\ TH,\ TT.$ They are four equally likely outcomes, and only $HH$ satisfies both events, confirming the probability $1/4$.

A bag contains $3$ red counters and $2$ blue counters. The probability of red on the first draw is $3/5$. If we do not replace the counter and the first one was red, only $2$ red counters remain among $4$, so $P(\text{red then red})=\frac35\cdot\frac24=\frac3{10}.$ Multiplying $3/5$ by $3/5$ would give $9/25$, which is wrong because the first draw changes the second probability. When evidence changes the next probability, keep the conditional factor. The guide to Bayes’ theorem shows how to update probabilities systematically.