# What is the probability of $A$?

$P$ is probability. We have: $P(A) \ge \frac{2}{3}$, $P(B) \ge \frac{2}{3}$, $P(C) \ge \frac{2}{3}$ and $P(A \cap B \cap C)=0$. We have to find $P(A)$. How to do it? Of course we have $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B)- P(A \cap C)- P(B \cap C)$, but what next? Please, help me.

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Hints:

• If $\mathbb P(B)\geqslant\frac23$ and $\mathbb P(C)\geqslant\frac23$, then $\mathbb P(B\cap C)\geqslant$ $____$.
• If $\mathbb P(A)=\frac23+c$ with $c\gt0$ and $\mathbb P(B\cap C)\geqslant$ $____$, then $\mathbb P(A\cap(B\cap C))\geqslant$ $____$.
• But the hypothesis is that $\mathbb P(A\cap B\cap C)=0$, hence $\mathbb P(A)=\frac23$.
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First, observe that

$$P(X \cap Y) = 0 \text{ implies } P(X \cap Y^c) = P(X). \tag{1}$$

Then, bound $P(B \cap C)$ by $$P(B \cap C) = P(B) + P(C) - P(B \cup C) \geq 2\frac{2}{3}-1 = \frac{1}{3}.$$

Finally, apply $(1)$ to $P(A\cap (B \cap C)) = 0$ to get $$P(A) = P(A \cap (B \cap C)^c) \leq \frac{2}{3}.$$

Concluding, $P(A) = \frac{2}{3}$.

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