Show that $P(C)=P(A\cup B)P(C\mid A)-P(A\cap B)P(C'\mid A)$

Let $A,B,$ and $C$ be events defined in the samplespace $S$, where

• $(A\cup B\cup C)\cap D=\emptyset$ and $ D\neq\emptyset$
• $C \subset (A\cup B)$ and $C\cap A\cap B = A\cap B$
• $A\cap B\neq \emptyset$ and $A\cap C\neq \emptyset$ and $B\cap C\neq \emptyset$
• $P(C\mid A)=P(C\mid B)$

1. Draw the events in a Venn Diagram. Explain why $D$ and $E=A\cup B\cup C$ are disjoint.

I have scetched a Venn Diagram like so:

Explanation:

Since we are given that $(A\cup B\cup C)\cap D=\emptyset$, we know that $D$and $E$ are disjoint, since $D\cup E=\emptyset\Rightarrow D$ and $E$ are disjoint!

Now comes the part I'm having trouble with:

2. Show that $P(C)=P(A\cup B)P(C\mid A)-P(A\cap B)P(C'\mid A)$

I started off with

\begin{align*} & 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) \\ & P(C) = P(A\cup B\cup C)-P(A)-P(B)+P(A\cap B)+P(A\cap C)+P(B\cap C)-P(A\cap B\cap C) \\ & P(C) = P(A\cup B\cup C)-P(A)-P(B)+P(A\cap B)+P(A\cap C)+P(B\cap C)-P(A\cap B) \\ \end{align*} (since we're given that $P(C\cap A\cap B) = P(A\cap B))$)

\begin{align*} & P(C) = P(A\cup B\cup C)-P(A)-P(B)+\(\cancel{P(A\cap B)}\)+P(A\cap C)+P(B\cap C)-\(\cancel{P(A\cap B)}\) \\ & P(C) = P(A\cup B\cup C)-P(A)-P(B)+P(A)+P(B)\\ & P(C) = P(A\cup B\cup C) \\ \end{align*}

This is where I got stuck, then I thought I'd try something different:

\begin{align*} & P(C) = P(A\cup B\cup C)-P(A)-P(B)+P(A\cap B)+P(A\cap C)+P(B\cap C)-P(A\cap B) \\ & P(C) = P(A\cup B\cup C)-P(A)-P(B)+P(A\cap B)+P(A)+P(B)-P(A\cap B)\\ & P(C) = P(A\cup B\cup C)-P(A)-P(B)+P(A\cap B)+P(A\cup B) \\ \end{align*}

But I got stuck here as well.

I tried to look at this question, but I didn't understand how to apply it to this exercise.

How do I prove this? All help is appreciated!

Solution 1:

I believe you have interpreted the second bullet point wrong. As you have drawn it, there are points in $A\cap B$ which are not in $C$ (thus not in $A\cap B\cap C$). From this bullet point, it must be the case that $(A\cap B)\subset C\subset (A\cup B)$, that is, $C$ contains the intersection of $A$ and $B$ rather than being contained by it.

From this, we get the following properties:

$$\color{green}{P(C|A\cap B)=1}$$ $$\color{red}{P(C|A'\cap B')=0}$$

Now to your question, note:

$$P(C)=P(C|A)P(A)+P(C|A')P(A')$$ $$=P(C|A)P(A)+P(C|A'\cap B)P(A'\cap B)+{\color{red}{P(C|A'\cap B')}}P(A'\cap B')$$ $$=P(C|A)P(A)+P(C|A'\cap B)P(A'\cap B)+{\color{red}{0}}$$ $$=P(C|A)P(A)+P(C|A'\cap B)P(A'\cap B)+P(C|A\cap B)P(A\cap B)-P(C|A\cap B)P(A\cap B)$$ $$=P(C|A)P(A)+P(C|B)P(B)-\color{green}{P(C|A\cap B)}P(A\cap B)$$ $$=P(C|A)(P(A)+P(B))-\color{green}{1}P(A\cap B)$$ $$=P(C|A)[P(A\cup B)+P(A\cap B)]-P(A\cap B)$$ $$=P(C|A)P(A\cup B)+[P(C|A)-1]P(A\cap B)$$ $$=P(C|A)P(A\cup B)-P(C'|A)P(A\cap B)$$