If $P(A \cup B \cup C) = 1$, $P(B) = 2P(A) $, $P(C) = 3P(A) $, $P(A \cap B) = P(A \cap C) = P(B \cap C) $, then $P(A) \le \frac14$ 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?

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 $P(A)\le \frac{1}{4}$.
