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Thread: conditional probability

  1. #1
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    conditional probability

    2 different numbers are chosen from the whole numbers from 1 to 9 inclusive.
    if one of the numbers is 5 what is the probability of the sum of the two numbers being equal

    so p(sum being even | getting 5) = probability of sum being even and getting a 5 /probability of getting a 5 =
    probability of sum being even and getting a 5/ 1/9
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  2. #2
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    Re: conditional probability

    Assuming equal probability for picking any number. There is a $2\dfrac{1}{9}$ probability of getting a 5. There is $\dfrac{1}{9}\dfrac{4}{8}+\dfrac{4}{9}\dfrac{1}{8} $ probability of getting a 5 and another odd number.

    Conditional probability is:

    $\dfrac{\dfrac{8}{72}}{\dfrac{2}{9}}=\dfrac{1}{2}$
    Last edited by SlipEternal; Jun 26th 2017 at 02:38 AM.
    Thanks from markosheehan
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  3. #3
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    Re: conditional probability

    [QUOTE=markosheehan;923479]2 different numbers are chosen from the whole numbers from 1 to 9 inclusive.
    if one of the numbers is 5 what is the probability of the sum of the two numbers being even?
    Looking back at your coin question, I must ask if you appreciate the power the power of the conditional?
    If you are given that one of the two numbers is a five, then for the sum to be five then in order for the sum is even the other must be odd. Five being taken there are eight left, of those only four are odd.
    Thus, $\mathscr{P}(E|5)=\dfrac{4}{8}$
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