# Thread: Conditional Probability Independence Proof

1. ## Conditional Probability Independence Proof

I have a problem with a simple conditional probability proof. Been working through it for a couple hours but couldn't make any inroads.

${\text{given that A and B are independent }} \hfill \\P\left( {A,B} \right) = P\left( A \right)P(B) \hfill \\{\text{Prove that}} \hfill \\P\left( {A|B,C} \right) = P(A|C) \hfill \\$

2. ## Re: Conditional Probability Independence Proof

Originally Posted by iamclarence
${\text{given that A and B are independent }} \hfill \\P\left( {A,B} \right) = P\left( A \right)P(B) \hfill \\{\text{Prove that}} \hfill \\P\left( {A|B,C} \right) = P(A|C) \hfill \\$
There some unusual and problematic notation here,
It appears that you are using $P(A,B)$ as the usual $P(A\cap B)$.
Does $P(A|B,C)$ mean $P(A|(B\cap C))~?$

3. ## Re: Conditional Probability Independence Proof

Thanks for the reply Plato, you are right, to rephrase,

Given that A and B are independent $P\left( {A \cap B} \right) = P\left( A \right)P(B)$
Prove that $P\left( {A|B \cap C} \right) = P(A|C)$

4. ## Re: Conditional Probability Independence Proof

Originally Posted by iamclarence
Given that A and B are independent $P\left( {A \cap B} \right) = P\left( A \right)P(B)$
Prove that $P\left( {A|B \cap C} \right) = P(A|C)$
Toss a die. Let $A$ be the event that a prime number appears.
Let $B$ be the event that a number less than three appears.
Let $C$ be the event that an odd number appears.

Is it clear that $A~\&~B$ are independent?

What are $\mathcal{P}(A|C)=~?~\&~ \mathcal{P}(A|B\cap C)=~?$

5. ## Re: Conditional Probability Independence Proof

Originally Posted by Plato
Toss a die. Let $A$ be the event that a prime number appears.
Let $B$ be the event that a number less than three appears.
Let $C$ be the event that an odd number appears.

Is it clear that $A~\&~B$ are independent?

What are $\mathcal{P}(A|C)=~?~\&~ \mathcal{P}(A|B\cap C)=~?$
Actually A and B are not independent because knowing information about B (that the number <3) means that there is a 1/2 chance of getting a prime number

P(A|B) = 1/2

P(A|C) would be given that an odd number appears, whats the chance that the number is prime
P(A|B^C) would be, given that the number is less than 3 and its odd, what are the chance that its prime

6. ## Re: Conditional Probability Independence Proof

Originally Posted by iamclarence
Actually A and B are not independent because knowing information about B (that the number <3) means that there is a 1/2 chance of getting a prime number
P(A|B)= 1/2 CORRECT
P(A|C) would be given that an odd number appears, whats the chance that the number is prime
P(A|B^C) would be, given that the number is less than 3 and its odd, what are the chance that its prime
Of course they are independent:
$\mathcal{P}(A)=\frac{1}{2}=\mathcal{P}(A|B)=\frac{ \tfrac{1}{6}}{\tfrac{2}{6}}=\frac{1}{2}$

$\mathcal{P}(A|C)=\frac{2}{3}~\&~ \mathcal{P}(A|B\cap C)=0?$