# Discrete Math Probability Proving.

• Sep 25th 2011, 02:11 PM
blackZ
Discrete Math Probability Proving.
Suppose that

a) P(A ∩ B ∩ C) = P(A) P(B) P(C)
P(A*∩ B ∩ C) = P(A*) P(B) P(C)
P(A ∩ B ̅ ∩ C) = P(A) P(B*) P(C)
P(A ∩ B ∩ C) = P(A) P(B) P(C*)
Prove that A B and C are independent.

b) Suppose that A B and C are independent. Then so are A* B* and C.

Note: A* means bar under A. Same for B & C.

Thanks.
• Sep 25th 2011, 02:26 PM
Plato
Re: Discrete Math Probability Proving.
Quote:

Originally Posted by blackZ
a) P(A ∩ B ∩ C) = P(A) P(B) P(C)
P(A*∩ B ∩ C) = P(A*) P(B) P(C)
P(A ∩ B ̅ ∩ C) = P(A) P(B*) P(C)
P(A ∩ B ∩ C) = P(A) P(B) P(C*)
Prove that A B and C are independent.
Note: A* means bar under A. Same for B & C.

\displaystyle \begin{align*} P(A\cap B) &=P(A\cap B\cap C)+ P(A\cap B\cap \overline{C})\\ &= P(A)P(B)P(C)+P(A)P(B)P(\overline{C})\\ &=P(A)P(B)[P(C)+P(\overline{C})]\\ &=P(A)P(B)[1] \\ &=P(A)P(B)\end{align*}
So $\displaystyle A~\&~B$ are independent.

Now you have to do the other two cases on your own.
• Sep 25th 2011, 02:33 PM
blackZ
Re: Discrete Math Probability Proving.
For a) In order to prove B&C as independent, do I need to replace A & B to B & C as you did. And later add all of them.
Also what is [1]? Is it part of the proof?
• Sep 25th 2011, 02:40 PM
Plato
Re: Discrete Math Probability Proving.
Quote:

Originally Posted by blackZ
Also what is [1]? Is it part of the proof?

That is the number 1: $\displaystyle P(X)+P(\overline{X})=1$