
Independent events
I am given that $\displaystyle P(A) >0, P(B) >0, P(C) > 0$ (strictly greater) and
$\displaystyle P(B)P(C) = \frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} + \frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})}$
I am to deduce whether B and C could be independent.
I am thinking no.
In order for the right side to be equal to $\displaystyle P(B \cap C)$ this means at least
$\displaystyle
\frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})} = 0 $
Now, $\displaystyle P(C) \neq 0$ since it was given that $\displaystyle P(C) > 0$
$\displaystyle P ( A \cap B) \neq 0$, since this would make $\displaystyle \frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} = 0 $
Therefore $\displaystyle P(B \cap C^{C}) = 0$
However $\displaystyle P(B \cap C^{C}) = 0 $ implies $\displaystyle B$ and $\displaystyle C^{C}$ are not independent in this case because $\displaystyle P(B) > 0$ and $\displaystyle P(C^{C}) > 0 $( since $\displaystyle P(C) > 0$)
$\displaystyle B$ and $\displaystyle C^{C}$ not independent implies $\displaystyle B $and $\displaystyle C$ aren't independent.
Is this sound reasoning? I am skeptical because this based of the converse of a true statement.
Thank you,
Jame

Wait
$\displaystyle P(C^{C})$ is not necessarily greater than zero. ( i.e. $\displaystyle if P(C) = 1$)
Darn, well at least what I said still works if $\displaystyle P(C) \neq 1 $

Edit: I forgot that we were assuming that $\displaystyle P(A  B \cap C) \neq P(AB)$.
So the $\displaystyle P(C)$ is not 1 and the $\displaystyle P(C^{C})$ is not zero.

Did you try a few numbers to see whether there might be a counter example?
maybe drawing a Venn diagram and throwing in some numbers where P(BC)=P(B)P(C) might shed light one way or another.