This problem is from the book, "Introduction to Probability" by Hoel, Port and Stone. It is problem 34 on page 25.
Problem:
Let $A$ and $B$ denote two independent events. Prove that $A$ and $B^c$ are independent.
Answer:
To show that $A$ and $B^c$ are independent, we I attempt to show that:
$$ P(A \cap B^c ) = P(A)P(B^c) $$
Since $A$ and $B$ are independent, we have:
\begin{align*}
P(A \cap B ) &= P(A)P(B) \\
P( A \cup B^c ) &= P(A) + P(B^c) – P(A \cap B^c ) \\
P(A \cap B^c ) &= P( A \cup B^c ) – P(A) – P(B^c) \\
P(A \cap B^c ) &= P( A \cup B^c ) – P(A) – ( 1 – P(B) ) \\
P(A \cap B^c ) &= P( A \cup B^c ) – P(A) – 1 + P(B)
\end{align*}
Now, I feel stuck. How do I finish the proof?
Best Answer
$A=A\cap (B\cup B^c)=(A\cap B)\cup (A\cap B^c)$
implies $P(A)=P(A\cap B)+P(A\cap B^c)-P(A\cap B\cap B^c)$, we deduce that $P(A)=P(A\cap B)+P(A\cap B^c)$.
$P(A\cap B^c)=P(A)-P(A\cap B)=P(A)-P(A)P(B)=P(A)(1-P(B))=P(A)P(B^c).$