Suppose that events $A$, $B$ and $C$ satisfy $P(A \cap B \cap C) = 0$ and each of them has probability not smaller than $\frac{2}{3}$. Find $P(A)$. Suppose that events $A$, $B$, and $C$ satisfy $P(A \cap B \cap C) = 0$ and each of them has probability not smaller than $\dfrac{2}{3}$. Find $P(A)$.


I don't understand this statement:


  $A \cap B$, $B \cap C$ and $C \cap A$ are (pairwise) disjoint events. Therefore, the sum of their probabilities is the probability of their union which cannot be greater than $1$.


Why is the sum of the probability of these three pairwise disjoint events $\leq 1$? shouldn't the probability of their union be $0$ such that the sum of their probabilities is not greater than zero? Please explain thanks.


  Solution: From the inclusion-exclusion formula
  $1 \geq P(A \cup B \cup C) = P(A) + P(B) + P(C) − P(A \cap B) − P(B \cap C) − P(C \cap A)$, hence $P(A \cap B) + P(B \cap C) + P(C \cap A) \geq 1$. But $A \cap B$, $B \cap C$ and $C \cap A$ are (pairwise) disjoint events. Therefore the sum of their probabilities is the probability of their union which cannot be greater than $1$. This means that we have equality everywhere, in particular, $P(A) = \dfrac{2}{3}$.
$P(A\cap B)=P(A\cap B \cap C^{c})$ because $P(A\cap B \cap C)=0$. So $P(A\cap B) \leq P(C^{c})=1-P(C) \leq 1/3$. Hence $P(A)=P(A \cap B)+P(A \cap B^{c}) \leq 1/3 + P(A \cap B^{c}) \leq 1/3 +(1-P(B)) \leq 1/3+1/3=2/3$. Hence $P(A)\leq 2/3$ and $P(A) \geq 2/3$ so $P(A)=2/3$.
Simpler: 

Consider the complements $A'$, $B'$ and $C'$, of $A$, $B$ and $C$, then $P(A'\cup B'\cup C')=1$ while $P(A')\leqslant\frac13$, $P(B')\leqslant\frac13$, and $P(C')\leqslant\frac13$ since $P(A)\geqslant\frac23$, $P(B)\geqslant\frac23$, and $P(C)\geqslant\frac23$. 

Since $P(A'\cup B'\cup C')\leqslant P(A')+P(B')+P(C')$, this implies that $P(A')=P(B')=P(C')=\frac13$.

In particular, $P(A)=\frac23$.