# Thread: [SOLVED] Help Needed to Prove Probability Theorem

1. ## [SOLVED] Help Needed to Prove Probability Theorem

I'm doing fine in all the other probability stuff I'm doing, but I got stuck in proving the following theorem:

If $A$, $B$ and $C$ are any three events, then

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
Source: Probability and Statistical Inference, 7e, by Hogg and Tanis

Here is my work thus far:

$P(A\cup B\cup C)=P(A\cup(B\cup C))$

This implies that
$P(A\cup(B\cup C))=P(A)+P(B\cup C)-P(A\cap(B\cup C))$

Now,
$P(B\cup C)=P(B)+P(C)-P(B\cap C)$

So, substituting this into the equation I get:

$P(A\cup(B\cup C))=P(A)+P(B)+P(C)-P(B\cap C)-{\color{red}P(A\cap(B\cup C))}$

I have no clue how to break up the part in red. Instead of working it out for me, give me a hint or two to help me get started.

Thanks in advance for any help!!

--Chris

2. It is easy. Take what you have in red.
$P\left( {A \cap \left[ {B \cup C} \right]} \right) = P\left( {\left[ {A \cap B} \right] \cup \left[ {A \cap C} \right]} \right) = P\left( {\left[ {A \cap B} \right]} \right) + P\left( {\left[ {A \cap C} \right]} \right) - P(A \cap B \cap C)$

3. Originally Posted by Plato
It is easy. Take what you have in red.
$P\left( {A \cap \left[ {B \cup C} \right]} \right) = P\left( {\left[ {A \cap B} \right] \cup \left[ {A \cap C} \right]} \right) = P\left( {\left[ {A \cap B} \right]} \right) + P\left( {\left[ {A \cap C} \right]} \right) - P(A \cap B \cap C)$
Ahh...I didn't apply the intersection properly. Thank you very much. Its a lot clearer now!

--Chris