What i have tried so far: given P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

P(A) + P(B) - P(A ∩ B) = P(A ∪ B) = P(A) + P(B|A′)P(A′)

P(B) - P(A ∩ B) = P(B|A′)P(A′)

I dont know how to continue, maybe my approach is entirely wrong :c

  • $\begingroup$ Just keep in mind the definition of conditional probabilities: $P(B|A')=\frac{P(B\cup A')}{P(A')}$. $\endgroup$ Dec 5, 2021 at 14:34

1 Answer 1


You have the right idea, it just gets a bit sidetracked in the middle. Your result can be proved as follows:

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ $$=P(A)+[P(A\cap B)+P(A'\cap B)]-P(A\cap B)$$ $$=P(A)+P(A'\cap B)$$ $$=P(A)+\frac{P(A'\cap B)}{P(A')}P(A')$$ $$=P(A)+P(B\mid A')P(A')$$

  • $\begingroup$ I know its right, but cant seem to phrase why the P(A′∩B) =P(B∣A′)P(A′). Could you please explain how did you get there? $\endgroup$
    – Harsh Dua
    Dec 5, 2021 at 14:43
  • $\begingroup$ @HarshDua I added a line to my solution. It should be clear now from the definition of conditional probability? $\endgroup$
    – Angelica
    Dec 5, 2021 at 14:58

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