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