I was working on some random probability problems given in exams throughout US colleges, and came across this relatively simple problem that is giving me a bit of trouble.
Suppose that $A$, $B$, and $C$ are pairwise independent events such that $P(A) = P(B) = P(C)$ and $A \cap B \cap C = \emptyset$. What is the largest possible value for $P(A)$?
Since $3P(A)−3P(A)^2=P(A\cup B\cup C)$ and $P(A\cup B\cup C) = P(A)+P(B∪C)−P(A\cap(B∪C))\geq 3P(A)−P(A)^2−1$.
I found that $P(A)$ could be at most $\frac{1}{\sqrt{2}}$, but I cannot find a set-up in which this happens.
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.
$$P(A)=P(B)=P(C)=\frac12$$ $$P(A\cap B)=P(B\cap C)=P(C\cap A)=\frac14$$ $$P(A\cap B\cap C)=0$$