Q:

Probability of one event given the probability of two other events

Let $A$, $B$, and $C$ be events. Suppose $P(A) \ge .9$, $P(B) \ge .8$, and $P(A \cap B \cap C)=0$. 
Show that $P(C) \le .3$.
Now, I tried using the inclusion-exclusion principle to solve this, but I'm getting nowhere. Perhaps that is the correct way of starting, but I'm looking at it the wrong way? It's been a little white since I've worked with this, so I'm not sure I'm on the right track.
Also, is it correct that $P(A \cap B \cap C)=0$ means that the events $A$, $B$, and $C$ are disjoint (but not necessarily $A \cap C$, $A \cap B$, and $B \cap C$)? 
Any hints would be appreciated.
Thanks.

A:

Since $A\cap B$ and $C$ are disjoint, 
$$\Pr(A\cap B)+\Pr(C)\le 1.\tag{$1$}$$
But $\Pr(A\cup B)\le 1$, so from the familiar 
$$\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$$
 we conclude that $\Pr(A\cap B)\ge 1.7-1=0.7$. 
Since $\Pr(A\cap B)\ge 0.7$, it follows from $(1)$ that $\Pr(C)\le 0.3$.
Remark: One can also use a formula-free argument, using a Venn diagram. That is better, but drawing and uploading a picture is (for me) not easy.

