Probability- Show that $ 1/6 \le P(A) \le 1/4$

Question

I have $A ∪ B ∪ C = Ω, P(B) = 2P(A), P(C) = 3P(A), P(A ∩ B) = P(B ∩ C)$. Pove that $$ 1/6 \le P(A) \le 1/4$$

Answer 1

Hint: Inclusion-exclusion states for three sets that

$Pr(A\cup B\cup C)=Pr(A)+Pr(B)+Pr(C)-Pr(A\cap B)-Pr(A\cap C)-Pr(B\cap C)+Pr(A\cap B\cap C)$

Using what we are told in the problem statement this simplifies:

$1=6Pr(A)-2Pr(A\cap B)-Pr(A\cap C)+Pr(A\cap B\cap C)$

Now, we can further manipulate this using what we know about the relationship between what still appears and axioms of probability such as $0\leq Pr(E)\leq 1$ and if $E\subseteq F$ then $Pr(E)\leq Pr(F)$.