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Math Help - Conditional Probability Independence Proof

  1. #1
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    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 \\
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    Re: Conditional Probability Independence Proof

    Quote Originally Posted by iamclarence View Post
    {\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))~?
    Last edited by Plato; October 19th 2011 at 05:28 AM.
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    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)
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    Re: Conditional Probability Independence Proof

    Quote Originally Posted by iamclarence View Post
    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)=~?
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    Re: Conditional Probability Independence Proof

    Quote Originally Posted by Plato View Post
    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
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    Re: Conditional Probability Independence Proof

    Quote Originally Posted by iamclarence View Post
    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?
    What does that tell you about this question?
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