# Independent events

• Feb 28th 2011, 05:42 PM
Jame
Independent events
I am given that $P(A) >0, P(B) >0, P(C) > 0$ (strictly greater) and

$P(B)P(C) = \frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} + \frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})}$

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 $P(B \cap C)$ this means at least

$
\frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})} = 0$

Now, $P(C) \neq 0$ since it was given that $P(C) > 0$
$P ( A \cap B) \neq 0$, since this would make $\frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} = 0$

Therefore $P(B \cap C^{C}) = 0$

However $P(B \cap C^{C}) = 0$ implies $B$ and $C^{C}$ are not independent in this case because $P(B) > 0$ and $P(C^{C}) > 0$( since $P(C) > 0$)

$B$ and $C^{C}$ not independent implies $B$and $C$ aren't independent.

Is this sound reasoning? I am skeptical because this based of the converse of a true statement.

Thank you,

-Jame
• Feb 28th 2011, 06:29 PM
Jame
Wait

$P(C^{C})$ is not necessarily greater than zero. ( i.e. $if P(C) = 1$)

Darn, well at least what I said still works if $P(C) \neq 1$
• Feb 28th 2011, 06:40 PM
Jame
Edit: I forgot that we were assuming that $P(A | B \cap C) \neq P(A|B)$.

So the $P(C)$ is not 1 and the $P(C^{C})$ is not zero.
• Mar 1st 2011, 01:06 PM
matheagle
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.