# Probability Question Proving?

• Oct 10th 2011, 08:18 PM
blackZ
Probability Question Proving?
Suppose that A, B and C are independent events and P(C) ≠ 0. Prove:
a) P (A ∩ B | C) = P(A | C) P(B | C)
b) P (A ∪ B | C) = P(A | C) + P(B | C) - P (A ∩ B | C)

My book only this formula to proved the above question.
p(A | B) =(A ∩ B)/ P(B)

I dont know how to deal with 3 constants?
Any help?

Thanks
• Oct 10th 2011, 08:43 PM
Soroban
Re: Probability Question Proving?
Hello, blackZ!

I'll do the first one . . .

Quote:

$\displaystyle \text{Suppose that }A, B, C\text{ are independent events and }P(C) \ne 0.$

$\displaystyle \text{Prove: }\:a)\;P(A\cap B\,|\,C) \:=\: P(A\,|\,C)\cdot P(B\,|\,C)$

Since $\displaystyle A,B,C$ are independent,
. . we have, for example: $\displaystyle P(A \cap B) \:=\:P(A)\cdot P(B)$

On the left side we have:

. . $\displaystyle P(A \cap B\,|\,C) \;=\;\frac{P(A \cap B \cap C)}{P(C)}$

. . . . . . . . $\displaystyle =\;\frac{P(A)\cdot P(B)\cdot P(C)}{P(C)} \;=\;P(A)\cdot P(B)$

On the right side we have:

. . $\displaystyle P(A\,|\,C)\cdot P(B\,|\,C) \;=\;\frac{P(A\cap C)}{P(C)}\cdot \frac{P(B\cap C)}{P(C)}$

. . . . . . . . $\displaystyle =\;\frac{P(A)\cdot P(C)}{P(C)} \cdot\frac{P(B)\cdot P(C)}{P(C)} \;=\;P(A)\cdot P(B)$

$\displaystyle \text{ta-}DAA!$

• Oct 11th 2011, 07:15 AM
blackZ
Re: Probability Question Proving?
Thanks for the (a).
For (b), I simplified the right side to P(A)+P(B) - P(A)P(B)

I am not sure what UNION and CONDITIONAL probability means when A,B and C are independent? Is it the same as INTERSECTION and CONDITIONAL?

Thanks
• Oct 11th 2011, 07:52 AM
Plato
Re: Probability Question Proving?
Quote:

Originally Posted by blackZ
Thanks for the (a).
For (b), I simplified the right side to P(A)+P(B) - P(A)P(B)
I am not sure what UNION and CONDITIONAL probability means when A,B and C are independent? Is it the same as INTERSECTION and CONDITIONAL?

\displaystyle \begin{align*}\mathcal{P}[(A\cup B)\cap C]&= \mathcal{P}[(A\cap C)+(B\cap C)]\\&= \mathcal{P}(A\cap C)+ \mathcal{P}(B\cap C)- \mathcal{P}(A\cap B\cap C)\\&= \mathcal{P}(A)\mathcal{P}(C)+ \mathcal{P}(B)\mathcal{P}(C) -\mathcal{P}(A) \mathcal{P}(B)\mathcal{P}(C)\\&= [\mathcal{P}(A)+ \mathcal{P}(B) -\mathcal{P}(A) \mathcal{P}(B)]\mathcal{P}(C)\\&= [\mathcal{P}(A\cup B)]\mathcal{P}(C) \end{align*}
• Oct 11th 2011, 07:55 AM
blackZ
Re: Probability Question Proving?
Quote:

Originally Posted by Plato
\displaystyle \begin{align*}\mathcal{P}[(A\cup B)\cap C]&= \mathcal{P}[(A\cap C)+(B\cap C)]\\&= \mathcal{P}(A\cap C)+ \mathcal{P}(B\cap C)- \mathcal{P}(A\cap B\cap C)\\&= \mathcal{P}(A)\mathcal{P}(C)+ \mathcal{P}(B)\mathcal{P}(C) -\mathcal{P}(A) \mathcal{P}(B)\mathcal{P}(C)\\&= [\mathcal{P}(A)+ \mathcal{P}(B) -\mathcal{P}(A) \mathcal{P}(B)]\mathcal{P}(C)\\&= [\mathcal{P}(A\cup B)]\mathcal{P}(C) \end{align*}

Thats more confusing. I dont know how the conditional probability | will go in there.
• Oct 11th 2011, 08:06 AM
Soroban
Re: Probability Question Proving?
Hello again, blackZ!

Quote:

For (b), I simplified the right side to: $\displaystyle P(A)+P(B) - P(A)\!\cdot\!P(B)$ . Good!

The left side is: .$\displaystyle P(A \cup B\,|\,C) \;=\;\frac{P\big([A \cup B] \cap C\big)}{P(C)}$ .[1]

We know that:
. . $\displaystyle P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)$
. x . . . . . . . $\displaystyle =\; P(A) + P(B) - P(A)\!\cdot\!P(B)$

Then [1] becomes:

.$\displaystyle \frac{\big[P(A) + P(B) - P(A)\!\cdot\!P(B)\big]\cdot P(C)}{P(C)} \;=\; P(A) + P(B) - P(A)\!\cdot\!P(B)$

• Oct 11th 2011, 08:13 AM
Plato
Re: Probability Question Proving?
Quote:

Originally Posted by blackZ
Thats more confusing. I dont know how the conditional probability | will go in there.

OK, I thought you could work through it.
\displaystyle \begin{align*}\mathcal{P}[(A\cup B)\cap C]&= \mathcal{P}[(A\cap C)+(B\cap C)]\\&= \mathcal{P}(A\cap C)+ \mathcal{P}(B\cap C)- \mathcal{P}(A\cap B\cap C)\\&= \mathcal{P}(A|C)\mathcal{P}(C)+ \mathcal{P}(B|C)\mathcal{P}(C) -\mathcal{P}(A\cap B|C)\mathcal{P}(C) \end{align*}

Divide by both sides by $\displaystyle \mathcal{P}(C)$