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We have ($P$ is probability): $P(A \cup B \cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A \cap B) = P(A \cap C) = P(B \cap C) $. Prove that $P(A) \le \frac{1}{4} $. Well, I tried with the fact that $ 1 = P(A \cup B \cup C) = 6P(A) - 3P(A \cap B) + P(A \cap B \cap C) $ but I got stuck... Could anyone help me, please?

Did
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Anne
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2 Answers2

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Let $\mathbb P(A)=a$, then $\mathbb P(B)=2a$ and $\mathbb P(C)=3a$. Hence $\mathbb P(B\cap C)\geqslant \mathbb P(B)+\mathbb P(C)-1$, that is, $\mathbb P(B\cap C)\geqslant5a-1$. By hypothesis, $\mathbb P(A\cap B)=\mathbb P(B\cap C)$ hence $\mathbb P(A\cap B)\geqslant5a-1$. But $\mathbb P(A\cap B)\leqslant \mathbb P(A)=a$ hence $a\geqslant5a-1$, that is, $\mathbb P(A)=a\leqslant\frac14$.

Note: This does not use the hypotheses on $\mathbb P(A\cup B\cup C)$ and $\mathbb P(A\cap C)$.

Did
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This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=P\big((A\cap B)\setminus C\big)$ and $y=P(A\cap B\cap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $A\setminus(B\cup C)$, $B\setminus(A\cup C)$, and $C\setminus(A\cup B)$, respectively. (A Venn diagram is helpful here.)

Now $$b+2x+y=2(a+2x+y)=2a+4x+2y\;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y\;,$$ so $c=3a+4x+2y$. Then

$$\begin{align*} 1&=P(A\cup B\cup C)\\ &=a+b+c+3x+y\\ &=6a+9x+4y\\ &=4(a+2x+y)+2a+x\\ &=4P(A)+2a+x\;. \end{align*}$$

Since $2a+x\ge 0$, we must have $P(A)\le \frac{1}{4}$.

Brian M. Scott
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  • But I guess $a= P(A) -2x+y$ and it changes a lot... – Anne Feb 19 '13 at 20:58
  • @Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(A\cap B)\setminus C$, as it was intended to be. – Brian M. Scott Feb 20 '13 at 00:29