Independent Events

Two events $A$ and $B$ are called independent if the occurrence of either one does not change the probability of the other - knowing one has happened gives no information about the other. Formally, $A$ and $B$ are independent when $P(A/B)=P(A/B')=P(A)\qquad\text{and}\qquad P(B/A)=P(B/A')=P(B).$ Combining this with the multiplication theorem $P(A\cap B)=P(A)\cdot P(B/A)$: for independent events, since $P(B/A)=P(B)$, we get the simple product rule $P(A\cap B)=P(A)\cdot P(B).$ More generally, if $A_1,A_2,\ldots,A_n$ are mutually independent, $P(A_1\cap A_2\cap\cdots\cap A_n)=P(A_1)P(A_2)\cdots P(A_n)$.

Theorem. If $A$ and $B$ are independent, then (a) $A$ and $B'$ are also independent, and (b) $A'$ and $B'$ are also independent.

Proof of (a): Using property 8, $P(A\cap B')=P(A)-P(A\cap B)=P(A)-P(A)P(B)=P(A)[1-P(B)]=P(A)\cdot P(B')$, which is exactly the product-rule signature of independence, so $A$ and $B'$ are independent.

Proof of (b): By De Morgan's law, $P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)=1-[P(A)+P(B)-P(A\cap B)]=1-P(A)-P(B)+P(A)P(B)=[1-P(A)]-P(B)[1-P(A)]=[1-P(A)][1-P(B)]=P(A')\cdot P(B')$, so $A'$ and $B'$ are independent too. …