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.

I've no idea how to begin with, so please help me out.

Thanks.

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.

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?

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**: $\displaystyle P(X)+P(\overline{X})=1$