MHB Solve Probability Inequality | Finite Probabilistic Space

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To solve the probability inequality Pr[A∩B] ≥ 9*Pr[A]*Pr[B] > 0, a finite probabilistic space can be defined with Ω = {a_1, a_2, ..., a_m} and uniform probabilities p(a_i) = 1/m for i = 1, 2, ..., m. By selecting events A = {a_1, a_2} and B = {a_1, a_3}, the condition simplifies to requiring m ≥ 36. This ensures that the intersection probability meets the inequality criteria. Thus, a finite space with at least 36 elements satisfies the problem's requirements.
alfred2
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Hi! I have to do this exercise:

Define a finite probabilistic Space (Ω; Pr[ ]) and 2 events A,B⊆ Ω and Pr[A] ≠ Pr so that we can verify that
Pr[A∩B]>=9*Pr[A]*Pr > 0. (1)
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I've been trying it but i have reached this conclusion:
If Pr[A]>0 Pr[A]=Pr[A∩B]/P[B|A]
IF Pr>0 Pr=Pr[A∩B]/P[A|B]
Substituting in (1) we have:
P[A|B]*P[A|B]>=9*Pr[A∩B]
I don't know if this help. Can anyone help me please? Thanks
 
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I solve the above question from an unanswered MHB thread.

Consider $\Omega=\{a_1,a_2,\ldots,a_m\}$ and the probabilty on $\mathcal{P}(\Omega)$ defined by: $$p(a_i)=\frac{1}{m}\quad \forall i=1,2,\ldots,m$$ Consider $A=\{a_1,a_2\}$ and $B=\{a_1,a_3\}$. Then, $$p(A\cap B)\ge 9p(A)p(B)\Leftrightarrow \frac{1}{m}\geq 9\cdot \frac{2}{m}\cdot\frac{2}{m}\Leftrightarrow 1\ge \frac{36}{m}\Leftrightarrow m\ge 36$$ That is, you only need to choose $\Omega$ with 36 elements.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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