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Please help me to prove that if $P(A^c∩B)=P(A^c)P(B)$ then $P(A∩B)=P(A)P(B)$.

This is what I have now:

$$P(A∩B) = P(B) - P(A^c∩B)$$ $$P(A∩B) = P(B) - P(A^c)P(B)$$ $$P(A∩B) = P(B)(1-P(A^c))$$

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  • $\begingroup$ Could you show us what you've tried? $\endgroup$
    – Jay Zha
    May 4, 2017 at 1:09
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    $\begingroup$ Your proof is complete! Because $1-P(A^c)$ is equal to $P(A)$. $\endgroup$
    – OmG
    May 4, 2017 at 1:12
  • $\begingroup$ @OmG By what law? $\endgroup$
    – socrates
    May 4, 2017 at 1:12
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    $\begingroup$ PS: It is the Additive Rule: "The probability measure for a union of disjoint events equals the sum of the probability measures for those events. " $\endgroup$ May 4, 2017 at 1:29
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    $\begingroup$ @OmG Not independent; complementary events are very dependent (if we know that one occured, then we are certain the other did not; by definition). What is important is that $A$ and $A^\complement$ are disjoint (also known as mutually exclusive). $\endgroup$ May 4, 2017 at 1:30

1 Answer 1

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Yes, that is okay.   The final step follows from the definition of complements.   $A\cup A^\complement$ is the sample space, and the probability measure for the sample space is unity.

$$\def\P{\operatorname{\mathsf P}} \begin{align}\P(A^\complement\cap B) &=\P(A^\complement)\P(B) &~(1)~& \text {Premise}\\[1ex]\P(A\cap B)+\P(A^\complement\cap B) &= \P(B) &~(2)~& \text{Law of Total Probability: a.k.a. the additive rule}\\[1ex] \P(A^\complement)+P(A) ~&=~ 1 &~(3)~& \text{same, definition of complement}\\[3ex] \P(A\cap B)&= \P(B)-\P(A^\complement\cap B) && \text{from } (2)\\[1ex] &= (1-\P(A^\complement))\P(B) && \text{from }(1)\\[1ex] &= \P(A)\P(B) && \text{from }(3) \end{align}$$

So if $\P(A^\complement\cap B)=\P(A^\complement)\P(B)$, then $\P(A\cap B)=\P(A)\P(B)$.

$\Box$

Similarly if $\P(A\cap B)=\P(A)\P(B)$, then $\P(A^\complement\cap B)=\P(A^\complement)\P(B)$.

$\blacksquare$

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