Simpson's paradox. Probability inequalities.

Question

I'm trying to solve this problem:

You have three Events: Event A: ill person gets well again. Event B: ill person takes medicine. Event C: ill person is male.

Now you have the following probabilities:

Show that: 1) P(A| B n C) > P(A| B^c n C )

2) P(A| B n C^c ) > P(A| B^c n C^c )

3) But P(A|B) = P(A| B^c )

I know that we somehow have to use the probabilities above, but I don't know how.

Thank you for you time.

Answer 1

Observe that the events $A\cap B\cap C, A\cap B\cap C^{\complement},\dots$ are $8$ sets that form a partition of the whole space $\Omega$. So the probability of every event that can be written as a union of these events (there are $2^8$ such unions) is determined.

I will just pick one out to illustrate.

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=$$$$\frac{P(A\cap B\cap C)+P(A\cap B\cap C^{\complement})}{P(A\cap B\cap C)+P(A\cap B\cap C^{\complement})+P(A^{\complement}\cap B\cap C)+P(A^{\complement}\cap B\cap C^{\complement})}$$

and leave the rest to you.