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Here is my question: I know if A & B are independent, P(A|B) = P(A); But If A and B are independent, is it true that: P(A'|B) = P (A')???

Amin
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  • what does A' mean? – Mr.Gandalf Sauron Oct 22 '21 at 11:15
  • Hint: $P(A|B) + P(A'|B) = 1$ – A. Webb Oct 22 '21 at 11:18
  • @Mr.GandalfSauron My guess, which could be mistaken, is that $A'$ represents the complement of $A$. Assuming so, the answer to the posted question is yes, if $A$ is independent of $B$, then $A'$ is independent of $B$. – user2661923 Oct 22 '21 at 11:19
  • @Mr.GandalfSauron Complement of A – Amin Oct 22 '21 at 11:19
  • The question has already been proven on mathSE. You might search for conditional probability, or other tags. I didn't want to go to the trouble of re-creating the proof, since I remember the issue already being broached on mathSE. I am unsure what mathSE posting has this issue, however. – user2661923 Oct 22 '21 at 11:22
  • @A.Webb You're Right! I ain't so good in math; so THAT was the key you just mentioned. Thanks a lot! – Amin Oct 22 '21 at 11:23
  • @user2661923 tnx a lot – Amin Oct 22 '21 at 11:24
  • https://math.stackexchange.com/questions/1962698/prove-that-if-events-a-and-b-are-independent-then-the-complement-events-of-a-an/2922539 This has been proven on stack exchange. How do I label this question a duplicate? – Gwendolyn Anderson Oct 22 '21 at 17:36

2 Answers2

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I know this question might've looked weird; But believe me, I don't like this subject much. That's why I couldn't solve such a simple question. Anyway, as @A.Webb commented: $P(A'|B) + P(A|B) = 1$, which could be written like this: $P(A'|B) = 1 - P(A|B)$.

As we had in hypothesis, A and B are independent, so $P(A|B) = P(A)$. Hence: $$P(A'|B) = 1 - P(A)$$ And we know $A$ and $A'$ are complements and to sum these two up, equals to $1$. Then $P(A'|B)$ MUST be equal to $P(A')$.

Siong Thye Goh
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Amin
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Proof

$$P(A'|B) =\frac{P(A' \cap B)}{P(B)}$$ But $P(A' \cap B) = P(A) - P(A n B)$ Thus $$P(A'| B) = P(B) \frac{1 - P(A)}{P(B)} = P(A')$$ since $1 - P(A) = P(A')$.

Siong Thye Goh
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Kidegalize
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