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I'm trying to solve this question and reach a proof, I have already proven that $P(B|A)>P(B)$, and $P(B^c|A)<P(B^c)$, using the information from the question, but I am struggling to find out how to prove that: $P(B^c|A^c)>P(B^c)$.

What I have Tried: I tried to write $(A^c \cap B^c)$ as $(A\cup B)^c$, and here's what I have reached:
$P(B^c|A^c)=\frac {P((A \cup B)^c)}{P(A^c)}=\frac {1-(P(A)+P(B)-P(A\cap B))}{1-P(A)}$, so in order to do my proof, I tried to substitute this into the inequality I'm trying to proof:
$\frac {1-(P(A)+P(B)-P(A\cap B))}{1-P(A)}-P(B^c)$, and started trying to proof that it is $>0$ (So I can add $P(B^c)$ and complete my proof).
But stuff got really messy and I couldn't really reach a point where I can say its bigger than zero.

I would really appreciate any feedback and help

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    $\begingroup$ hello, in the question title, you are calculating $P(A^c|B^c)$ but in the second line of "what I tried" you are calculating $P(B^c|A^c)$. just want to ensure that the question in the title is what we want to prove? $\endgroup$ Apr 14, 2021 at 15:40
  • $\begingroup$ @RahulMadhavan Thanks I made a mistake in the title, edited it now $\endgroup$
    – Pwaol
    Apr 14, 2021 at 15:43

2 Answers 2

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You have said that you already showed $P(B|A)>P(B)$. I will use this. \begin{align*} P(B|A)&>P(B)\\ \frac{P(B\cap A)}{P(A)}&>P(B)\\ P(B\cap A)&>P(B)P(A)\tag 1 \end{align*} Now consider $P(A^C)P(B^C)$ \begin{align*} P(A^C)P(B^C) &=(1-P(A))(1-P(B))\\ &=1-P(A)-P(B)+P(A)P(B)\\ &<1-P(A)-P(B)+P(A\cap B)\tag{By 1}\\ &=1-(P(A)+P(B)-P(A\cap B))\\ &=1-P(A\cup B)\\ &=P\left((A\cup B)^C\right)\\ &=P\left(A^C\cap B^C\right)\tag{by DeMorgan's laws}\\ \end{align*} Therefore: \begin{align*} P(A^C)P(B^C)&<P\left(A^C\cap B^C\right)\\ P(B^C)&<\frac{P\left(A^C\cap B^C\right)}{P(A^C)}\\ P(B^C)&<P\left(B^C|A^C\right)\\ \end{align*}

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  • $\begingroup$ Appreciate the detailed answer, I got a general question if possible, usually in questions like these, do I need to worry about if $P(B)=0$, or if $P(A^c)=0$? I completely ignored these possibilities and wondering if it's usually taken for granted in probability or not $\endgroup$
    – Pwaol
    Apr 14, 2021 at 19:26
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    $\begingroup$ Thank Pwaol. Yes, you should worry about these cases. Conditioning on null events is not defined. see here: math.stackexchange.com/questions/1903652/… $\endgroup$ Apr 14, 2021 at 19:31
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A solution could be the following:

$$P(B^c | A^c) = P(A^c \cap B^c)/P(A^c) = [1-(P(A)+P(B)-P(A \cap B))]/P(A^c) $$

Now we observe that the above value is bigger than:

$$[1-P(A) - P(B) + P(A)\times P(B)]/P(A^c)$$

since $P(B | A) > P(A)\times P(B)$.

Perceive that I did $1-P(A) = P(A^c) $ above and factored $P(B)[P(A) -1] = P(B)\times -P(A^c)$

And so,

$P(B^c|A^c) > P(B^c)$.

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