0
$\begingroup$

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!

$\endgroup$

closed as off-topic by user99914, Namaste, Did, The Phenotype, NCh Jan 29 '18 at 0:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Namaste, Did, The Phenotype, NCh
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ what did you try? I would try to use $P (A\cup B)-P (A \cap B)=P (A)+P (B)$ $\endgroup$ – user519338 Jan 28 '18 at 23:43
1
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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 $P(A)=\frac23$.


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 :)

$\endgroup$
  • $\begingroup$ +1 for general $n$. I didn't find your version of Bonferroni's inequality, can you give me a link to it? Thank you. $\endgroup$ – Weijun Zhou Jan 29 '18 at 0:08
  • $\begingroup$ Can't find a link but have added a proof in my answer. $\endgroup$ – David Jan 29 '18 at 0:42
  • $\begingroup$ Thank you very much. It is an elegant proof. $\endgroup$ – Weijun Zhou Jan 29 '18 at 0:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.