# Prove that if $P(A^c∩B)=P(A^c)P(B)$ then $P(A∩B)=P(A)P(B)$.

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:

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

• Could you show us what you've tried? May 4, 2017 at 1:09
• Your proof is complete! Because $1-P(A^c)$ is equal to $P(A)$.
– OmG
May 4, 2017 at 1:12
• @OmG By what law? May 4, 2017 at 1:12
• 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. " May 4, 2017 at 1:29
• @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). May 4, 2017 at 1:30

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 $\P(A^\complement\cap B)=\P(A^\complement)\P(B)$.
$\blacksquare$