Maximum possible probability given null interesection of events
Solution 1:
Let $a=P(A)$.
$a=P(A)$
$\geq P(A\cap (B\cup C))$
$=P(A\cap B)+P(A\cap C) - P(A\cap B\cap C)$
$=P(A) P(B)+P(A) P(C)$
$=2a^2$
$a\geq 2a^2$
$a\leq \frac12$
construction:
Let there be 3 fair coins. Flip all of them. Let A be the event that coin flips 1 and 2 have different outcomes. Let B be the event that coin flips 2 and 3 have different outcomes. Let C be the event that coin flips 3 and 1 have different outcomes. Then all the events cannot hold simultaneously.
$$P(A)=P(B)=P(C)=\frac12$$ $$P(A\cap B)=P(B\cap C)=P(C\cap A)=\frac14$$ $$P(A\cap B\cap C)=0$$