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-   -   If A, B independent events. Then A and B^C, as well as, A^C and B are independent (http://mymathforum.com/probability-statistics/57922-if-b-independent-events-then-b-c-well-c-b-independent.html)

 beesee June 9th, 2015 09:10 AM

If A, B independent events. Then A and B^C, as well as, A^C and B are independent

Hi,

I want to prove the following, but I encounter some problems. Can you give me any suggestions?

Let A, B be independent events. Show that $\displaystyle A$ and $\displaystyle B^C$ are independent.

I have done some things but without success:
So, we have to show that $\displaystyle P(A \cap B^C) = P(A) \cdot P(B^C)$.
Let:
$\displaystyle P(A) = x \Rightarrow P(A^C) = 1 - P(A) = 1- x, \\ P(B) = y \Rightarrow P(B^C) = 1 - P(B) = 1 - y$

Since A, B are independent events we have:
$\displaystyle P(A \cap B) = P(A) \cdot P(B) = x \cdot y$

So here we want to know:
"In what way do we can rewrite $\displaystyle P(A \cap B^C)$, from the math data available?"

I have tried this, using DeMorgan, but without obtain what I needed:
$\displaystyle P[(A^C \cup B)^C] = P[(A^C)^C \cap B^C] = P(A \cap B^C) = 1 - P(A^C \cup B) =$
here I'm not sure if the addition rule can be applied, however:
$\displaystyle = 1 - [P(A^C) + P(B) - P(A^C \cap B)]$

but here I don't have any $\displaystyle P(A^C \cap B)$ term to substitute using x,y. So I don't know how to proceed!

Can you give any suggestions?
Many thanks!

 beesee June 9th, 2015 01:33 PM

for clarity I post here another proof, because it is from the following one that I have to prove what I have written in the first post:

Suppose $\displaystyle A$ and $\displaystyle B$ are independent events. Then $\displaystyle A^C$ and $\displaystyle B^C$ are independent events.

We have to show that $\displaystyle P(A^C \cap B^C) = P(A^C) \cdot P(B^C)$.

Let:
$\displaystyle P(A) = x \Rightarrow P(A^C) = 1 - P(A) = 1- x, \\ P(B) = y \Rightarrow P(B^C) = 1 - P(B) = 1 - y$

Since A, B are independent events we have:
$\displaystyle P(A \cap B) = P(A) \cdot P(B) = x \cdot y$

$\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = x + y - xy$

Using DeMorgan law:
$\displaystyle P(A^C \cap B^C) = P[(A \cup B)^C] = 1 - P(A \cup B) = 1 - x - y + xy.$

And also:
$\displaystyle P(A^C) \cdot P(B^C) = (1-x)(1-y) = 1-x-y+xy$

END.

Similarly, I have tried to show in the first post that A and B^C, as well as A^C and B are independent, but without success. Can you give me any helping hand? many thanks!

 mathman June 9th, 2015 01:52 PM

P(A)=P(A∩B)+P(A∩B')=P(A)P(B)+P(A∩B') (B and B' are disjoint and add to entire space)
P(A∩B')=P(A)(1-P(B))=P(A)P(B')

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