\problemdata      
        {probability-conditional-6} % latex-friendly label for the prob.
        {conditional probability} % The topic(s) of the problem content.
        {F00 ps10} % Source (if known)
        {S02 cp9F} % Usage (list ones you are aware of).
        {Albert Meyer, S02} % Last revision info (author, date).

\begin{problem}\label{ABC}
Let $A,B,C$ be events.  For each of the following statements, prove it or
give a counterexample.

\bparts

\ppart\label{ABCI} If $C$ is independent of $A$, and $C$ is independent of $B$, then
$C$ is independent of $A \intersect B$.

\solution{Let the sample space be the set of four length-two bit strings.
Define $A$ to be the event that the first bit is 1, $B$ to be the event
that the second bit is 1, and $C$ to be the event that at exactly one of
the bits is 1.

Note $A$, $B$ and $C$ each consist of two sample points, and the
intersection of any two of them consists of one point.  So each has
probability 2/4=1/2, while the intersection of any two of them has
probability $1/4$.  Therefore the product, $(1/2)\cdot(1/2)$, of the
probabilities of any two of these events equals the probability of their
intersection.  So any two among $A,B$ and $C$ are independent, \ie they are
\emph{pairwise} independent.

But there are no points in all three of $A$, $B$ and $C$.  So
\[
\pr{C \intersect (A \intersect B)} = 0 \neq \frac{1}{8} = \frac{1}{2}
\cdot \frac{1}{4} = \pr{C}\pr{A \intersect B}.
\]
Hence, $C$ is not independent of $A \union B$.
}

\ppart\label{ABOPLUS} If $C$ is independent of $A$, and $C$ is independent
of $B$, then $C$ is independent of $A \union B$.

\solution{
The statement is false.

The countexample from part~(\ref{ABCI}) works here too.

We already observed that $A$, $B$ and $C$ are pairwise independent.  We
claim that $C$ is not independent of the event $A \cup B$.  This follows
because $A \union B$ consists of the two points in $C$ as well as the point
\texttt{11}.  So
\[
\pr{C \intersect (A \union B)} = \pr{C} = \frac{1}{2} \neq \frac{1}{3} =
\frac{1}{2} \cdot \frac{2}{3} = \pr{C}\pr{A \union B}.
\]
}

\ppart If $C$ is independent of $A$, and $C$ is independent of $B$, and
$C$ is independent of $A \intersect B$, then $C$ is independent of $A
\union B$.

\solution{The statement is true:
\begin{align*}
\pr{C \intersect (A \union B)} & = \pr{(C \intersect A) \union (C \intersect B)}
          & \text{(distributivity of $\intersect$ over $\union$)}\\
  & =  \pr{C \intersect A} + \pr{C \intersect B} - \pr{(C \intersect A \intersect B)} 
          & \text{(inclusion-exclusion)}\\
  & = \pr{C}\pr{A} + \pr{C}\pr{B} - \pr{C}\pr{A \intersect B}
          & \text{(independence)}\\
  & = \pr{C}(\pr{A} + \pr{B} - \pr{A \intersect B}) & \text{(factoring)}\\
  & = \pr{C}(\pr{A \union B}) & \text{(inclusion-exclusion)}.
\end{align*}
}

\eparts

\end{problem}
