Help with a Probability Proof How do I prove the following:
Show that if $A$, $B$ and $C$ are three events such that
$P(A \cap B \cap C) \neq 0$ and $P(C | A \cap B)$ = $P(C|B)$,
then $P(A| B \cap C) = P(A | B)$.
Here is my attempt
$P(C|A \cap B) = { {P(A \cap B \cap C)} \over {P(A \cap B)}} = P(C|B) = {{P(C \cap B)} \over {P(B)}}$
Now, I do not know what to do. I am hoping somebody can help me. This is not a homework problem.
Bob
 A: Since $P(A\cap B)>0$ by hypothesis, one can do the following:
$$P(A | B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}\frac{P(A\cap B)}{P(A\cap B)}=$$
$$=\frac{P(A\cap B\cap C)}{P(A\cap B)}\frac{P(A\cap B)}{P(B\cap C)}=$$
$$=P(C | A\cap B)\frac{P(A\cap B)}{P(B\cap C)}.$$
Now, from hypothesis againg:
$$P(C | A\cap B)=P(C | B)=\frac{P(C\cap B)}{P(B)}.$$
That is,
$$P(A | B\cap C)=\frac{P(C\cap B)}{P(B)}\frac{P(A\cap B)}{P(B\cap C)}=$$
$$=\frac{P(A\cap B)}{P(B)}=P(A | B).$$
A: Hint.  Write the equation $P(A\mid B\cap C)=P(A\mid B)$ in terms of quotients of probabilities.  Then compare it carefully with the equation you have already written down in your question.  A little simple algebra is all that is needed.
A: You are very close, you just need to do a little rearrangement.
From where you're up to:
$$
\begin{eqnarray}\frac{P(A\cap B \cap C)}{P(A \cap B)} & = &\frac{P(C \cap B)}{P(B)}\\
\Rightarrow P(A\cap B \cap C)\cdot P(B) & = & P(C \cap B)\cdot P(A \cap B)\\
\Rightarrow \frac{P(B)}{P(A\cap B)} & = & \frac{P(C\cap B)}{P(A\cap B \cap C)}\\
\Rightarrow \frac{1}{P(A\mid B)} & = & \frac{1}{P(A\mid B \cap C)}\\
\Rightarrow P(A\mid B) & = & P(A\mid B \cap C)
\end{eqnarray}
$$
