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I am stuck. The problem is if A and B are independent events prove that A' and B' are also independent. I got P(A union B) must be 1/2. I am wrong?

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Yes, you are probably wrong. What is your definition of independence? What is $\Pr(A')$? –  Henry Apr 24 '12 at 22:45
    
My definition is P(A')=1-P(A) –  dot dot Apr 24 '12 at 22:45

3 Answers 3

up vote 1 down vote accepted

Assume $A$ and $B$ are independent. Then
$P(A \cap B) = P(A) P(B)$, or equivalently, $P(B) = P(B|A)$, $P(A) = P(A|B)$.

Playing off of the hint given in Ross Millikan's answer, here is another approach using conditional probability.

$$P(B'|A) = \frac{P(B'\cap A)}{P(A)} = \frac{P(A \setminus (B\cap A))}{P(A)} = \frac{P(A) - P(B\cap A)}{P(A)} = \frac{P(A) - P(B)P(A)}{P(A)}$$ $$ = 1 - P(B) = P(B')$$

This shows that $P(B'|A) = P(B')$, and hence $B'$ and $A$ are independent. We want to show that $P(B'|A') = P(B')$. $$P(B'|A') = \frac{P(B'\cap A')}{P(A')} =\frac{P(B'\setminus(B'\cap A))}{P(A')} = \frac{P(B') - P(B'\cap A)}{1 - P(A)}$$ and since we previously proved that $B'$ and $A$ are independent, $$= \frac{P(B') - P(B')P(A)}{1 - P(A)} = \frac{P(B')(1-P(A))}{1-P(A)} = P(B')$$

Hence we have proven that $A'$ and $B'$ are independent.

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Which assumes that $P(A)\ne0$ and $P(A)\ne1$. Since this hypothesis is unnecessary, this is a good indication that conditional probabilities should be avoided here. For a proof with no division, see @David's answer. –  Did Jun 19 '12 at 7:04

Since, by Demorgan's law, $ A'\cap B' =(A\cup B)'$

$$\eqalign{ P(A'\cap B')&= 1-P(A\cup B )\cr &=1-\bigl[ P(A)+P(B)-P(AB)\bigr]\cr &=1-P(A)-P(B)+P(AB)\cr &=1-P(A)-P(B)+P(A)P(B)\cr &=P(A')-P(B)\bigl(1-P(A)\bigr)\cr &=P(A')-P(B)P(A')\cr &=P(A')\bigl(1-P(B)\bigr)\cr &=P(A')P(B'). } $$

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You have no data to say what $P(A\cup B)$ is. $A$ could be rolling a $1$ on a red die and $B$ rolling a $1$ on a green die. Certainly $P(A\cup B)\neq \frac 12$

Independent just means that the occurrence of $A$ doesn't change the chance of $B$, so $P(B)=P(B|A)=P(B|A')$ You want to show that $P(B')=P(B'|A')$. It might help to make a Venn diagram and label the probabilities of each region.

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