Let A,B be two events such that $P(A|B) > P(B)$, Prove that $P(B^c|A^c)>P(B^c)$

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

• 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? Apr 14, 2021 at 15:40
• @RahulMadhavan Thanks I made a mistake in the title, edited it now Apr 14, 2021 at 15:43

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)&

• 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 Apr 14, 2021 at 19:26
• Thank Pwaol. Yes, you should worry about these cases. Conditioning on null events is not defined. see here: math.stackexchange.com/questions/1903652/… Apr 14, 2021 at 19:31

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,

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