\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-k-Events} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, TU11} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).


\begin{problem}
Let $A,B,C$ be events of some experiment.  For each of the following 
statements, prove it if it is true or give a counterexample if it is
false:

\bparts

\ppart 
If $A$ is independent of $B$, 
then $A$ is also independent of $\overline{B}$.

\solution{ 
This is true. To prove it, suppose $A$ is independent of $B$; that
is, $\prob{A\cap B}=\prob{A}\prob{B}$. Then
\begin{align}
\label{eq:pr-rule-1}
\prob{A\cap \overline{B}}
&=
\prob{A}\prob{\overline{B} \mid A}
\\
\label{eq:co-rule-1}
&=
\prob{A}(1-\prob{B\mid A})
\\
\nonumber
&=
\prob{A}-\prob{A}\prob{B\mid A}
\\
\label{eq:pr-rule-2}
&=
\prob{A}-\prob{A\cap B}
\\
\label{eq:independence}
&=
\prob{A}-\prob{A}\prob{B}
\\
\nonumber
&=
\prob{A}(1-\prob{B})
\\
\label{eq:co-rule-2}
&=
\prob{A}\prob{\overline{B}}
\end{align}
Equations~\eqref{eq:pr-rule-1} and~\eqref{eq:pr-rule-2} make use of 
the Product Rule. The Complement Rule is used in~\eqref{eq:co-rule-1}
and~\eqref{eq:co-rule-2}. We use independence of $A$ and $B$
in~\eqref{eq:independence}. 
}

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

\solution{
This is false. To see why, consider the usual random experiment of
rolling a die and let $A$, $B$, $C$ be the events that the die rolls
less than 3, rolls an even number, or rolls a prime number,
respectively. I.e., 
$$
A = \{1,2\} \qquad
B = \{2,4,6\} \qquad
C = \{2,3,5\}.
$$
Then, 
$$
\prob{A}=\tfrac{1}{3}, 
\qquad
\prob{B}=\prob{C}=\tfrac{1}{2},
$$
and
$$
\prob{A\cap B}=
\prob{A\cap C}=
\prob{B\cap C}=
\prob{A\cap B\cap C}=
\prob{\{2\}}=
\tfrac{1}{6}.
$$
Easily, 
$$
\prob{A\mid B}=
\prob{A\mid C}=
\tfrac{1}{6}/\tfrac{1}{2}=
\tfrac{1}{3}=
\prob{A},
$$
which proves $A$ is independent of $B$ and also independent of
$C$. However, 
$$
\prob{A\mid B\cap C}=
\prob{A\cap B\cap C}/\prob{B\cap C}=
\tfrac{1}{6}/\tfrac{1}{6}=
1\neq
\prob{A},
$$
which implies $A$ is not independent of $B\cap C$.
}

\ppart 
If $A$ is independent of $B$, and $A$ is independent of $C$, and $A$
is independent of $B\cap C$, 
then $A$ is independent of $B\cup C$.

\solution{
This is true. To prove it, suppose $A$ is independent of any one of
$B$, $C$, and $B\cap C$; that is, 
\begin{align*}
\prob{A\cap B} &= \prob{A}\prob{B} \\
\prob{A\cap C} &= \prob{A}\prob{C} \\
\prob{A\cap (B\cap C)} &= \prob{A}\prob{B\cap C}.
\end{align*}
Then, 
\begin{align}
\nonumber
\prob{A\cap(B\cup C)}
&=
\prob{(A\cap B)\cup (A\cap C)} 
\\
\label{eq:ie-rule-1}
&=
\prob{A\cap B}+\prob{A\cap C}-\prob{(A\cap B)\cap(A\cap C)}
\\
\nonumber
&=
\prob{A\cap B}+\prob{A\cap C}-\prob{A\cap B\cap C}
\\
\label{eq:indep}
&=
\prob{A}\prob{B}+\prob{A}\prob{C}-\prob{A}\prob{B\cap C}
\\
\nonumber
&=
\prob{A}(\prob{B}+\prob{C}-\prob{B\cap C})
\\
\label{eq:ie-rule-2}
&=
\prob{A}\prob{B\cup C}.
\end{align}
Equations~\eqref{eq:ie-rule-1} and~\eqref{eq:ie-rule-2} use the
Inclusion-Exclusion Rule. Equation~\eqref{eq:indep} uses the
independences of $A$.
}

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

\solution{
This is false. E.g., look again at the counter-example for part~(b)
and notice that  
$$
\prob{B\cup C}=\tfrac{5}{6}, \qquad
\prob{A\cap(B\cup C)}=\tfrac{1}{6}.
$$ 
Hence,
$$
\prob{A\mid B\cup C}=
\tfrac{1}{6}/\tfrac{5}{6}=
\tfrac{1}{5}\neq
\prob{A}
$$
which again implies $A$ is not idependent of $B\cup C$.
}

\eparts



\end{problem}