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Unit 01 /
Problem 2:


F computing probabilities with independent events
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Hard


Assume two events, $A$ and $B$ are independent ($A \perp B$). a) prove that $A^C$ and $B^C$ are independent. b) prove that $A$ and $B^C$ are independent.

Solution:

(a)\[\begin{align}P(A^C \cap B^C)&=1-P((A^C\cap B^C)^C),\text{ complement rule}\\
&=1-P(A\cup B),\text{ De Morgan's law}\\
&=1-[P(A)+P(B)-P(A\cap B)],\text{ addition rule}\\
&=1-[P(A)+P(B)-P(A)P(B)],\text{ using independence}\\
&=1-P(A)-P(B)+P(A)P(B)\\
&=(1-P(A))(1-P(B))\\
&=P(A^C)P(B^C),\text{ complement rule}\end{align}\]Hence, $A^C$ and $B^C$ are independent.

(b)We note first that $A$ is the union of the disjoint sets $(A\cap B^C)$ and $(A\cap B)$. By the addition rule, we then get\[\begin{align}&P(A)=P(A\cap B^C)+P(A\cap B)\\
\implies &P(A)=P(A\cap B^C)+P(A)P(B),\text{ since independent}\\
\implies &P(A \cap B^C)=P(A)-P(A)P(B)\\
\implies &P(A\cap B^C)=P(A)(1-P(B))\\
\implies &P(A\cap B^C)=P(A)P(B^C),\text{ complement rule}\end{align}\]Hence, $A$ and $B^C$ are independent. A similar argument will also show that $A^C$ and $B$ are independent.