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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: enter image description here

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.

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    $\begingroup$ You might want to draw a Venn diagram with an outer rectangle for the universal set and three circles for A, B, and C. It's then easy to do the computations mentioned in the answer below without writing all those hairy formulas. 😀 $\endgroup$
    – Ned
    Commented Nov 12, 2017 at 16:47

1 Answer 1

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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.

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  • $\begingroup$ Thank you so much! I did it! I was able to show the three (in)equalities! I had just one little problem: what if B and C are independent? Does the equality in 3) still hold? $\endgroup$
    – RukiaKuchiki
    Commented Nov 13, 2017 at 11:17
  • $\begingroup$ In 3) the events involved are $A,B$ (not $C$). $\endgroup$
    – drhab
    Commented Nov 13, 2017 at 11:24

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