# Probability Inequality

#### alfred

##### New member
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

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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.