# 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.

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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.

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Thanks very much for your answer. I was sure that the inclusion-exclusion principle was involved, just overlooked the step from equation (1) in your answer. I agree that a Venn diagram makes it quite clear. –  mamyot0 Oct 6 '12 at 2:12