Suppose that events $A$, $B$ and $C$ satisfy $P(A\cap B\cap C) = 0$

Question

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 $P(A)$.

I'm confused if there is a unique $P(A)$? Thanks!

Answer 1

$$P(A\cap B\cap C)=P(A)+P(B)+P(C)-[P(A\cup B)+P(B\cup C)+P(A\cup C)-P(A\cup B\cup C)]=0,$$ $$P(A\cup B)+P(B\cup C)+P(A\cup C)-P(A\cup B\cup C)\ge2,$$

but $$P(A\cup B)+P(B\cup C)+[P(A\cup C)-P(A\cup B\cup C)]\le 1+1+0=2,$$

so $$P(A\cup B)+P(B\cup C)+P(A\cup C)-P(A\cup B\cup C)=2,$$

this in turn results in, $$P(A)+P(B)+P(C)=2,$$

but, $$P(A),P(B),P(C)\ge\frac23,$$

so, $$P(A)=P(B)=P(C)=\frac23.$$

Answer 2

Bonferroni's inequality states that $$P(A_1\cap\cdots\cap A_n) \ge P(A_1)+\cdots+P(A_n)-(n-1)P(A_1\cup\cdots\cup A_n)\ .$$ Take $n=3$, relabel the events as $A,B,C$ and solve for $P(A)$ to give $$P(A)\le P(A\cap B\cap C)+2P(A\cup B\cup C)-P(B)-P(C)\ .$$ Now obviously $P(A\cup B\cup C)\le1$, it is given that $P(A\cap B\cap C)=0$ and it is also given that $P(B),P(C)\ge\frac23$. Substituting yields $$P(A)\le\frac23\ ,$$ and you already know $P(A)\ge\frac23$, hence $P(A)=\frac23$.

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Proof of the inequality: for sets $A_1,\ldots,A_n$, count every element of $A_1\cup\cdots\cup A_n$ once for each set which contains it. This gives $$|A_1|+\cdots+|A_n|\ .$$ However, every element has been counted at most $n-1$ times, except for the elements of the intersection which have been counted once more. So the count is at most $$(n-1)|A_1\cup\cdots\cup A_n|+|A_1\cap\cdots\cap A_n|\ ,$$ which gives the counting version of Bonferroni's inequality, and dividing by$P({\cal U})$ gives the version for (discrete) probability. Don't know of any equally elegant proof for continuous probability :)