Positive information of an event proof An event F is said to carry positive information about an event E ($F\to E$) if $P[E|F]\ge P[E]$
Prove or give counterexamples to the following assertions:
1)if $F\to E$ and $E\to G$, then $F\to G$
2)if $F\to E$ and $G\to E$, then $F\cap G\to E$
I tried to use bayes formula but I got stuck, and I can´t think of any counterexamples. Can you give me a hand with this problem? Iwould really appreciate it :) 
 A: A counterexample to assertion (1):  Let the table of probabilities of the three events e f and g be:
$$
\begin{array}{cccc} 
g & e & f & Probability \\
\hline 
F & F & F & 50\% \\
T & F & F & 10\% \\
F & T & F & 10\% \\
T & T & F & 10\% \\
F & F & T & 10\% \\
T & F & T & 0\% \\
F & T & T & 10\% \\
T & T & T & 0\% \\
\end{array}
$$
Then $P(e | f) = \frac{1}{2} > P(e) = \frac{3}{10}$ so $f \rightarrow e$.
And $P(g | e) = \frac{1}{3} > P(g) = \frac{1}{5}$ so $e \rightarrow g$.
But $P(g | f) = 0 < P(g) = \frac{1}{5}$ so $f \rightarrow g$ is false.
A counterexample to assertion (2) has the table
$$
\begin{array}{cccc} 
e & g & f & Probability \\
\hline 
F & F & F & 30\% \\
T & F & F & 5\% \\
F & T & F & 0\% \\
T & T & F & 25\% \\
F & F & T & 0\% \\
T & F & T & 25\% \\
F & T & T & 20\% \\
T & T & T & 0\% \\
\end{array}
$$
where $P(e) = 55\%$, $P(e|f) = P(e|g) = \frac{5}{9}$ so $f \rightarrow e$ and $g\rightarrow e$.  
Yet $P(e|(f \wedge g) = 0$ so $(f \wedge g) \rightarrow e$ is patently false.
