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Thread: Probability to draw an odd number before an even number

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    Probability to draw an odd number before an even number




    There's 39 numbers in a jar and we're picking them without putting them back. The probability of drawing an odd number and then an even number is (20/39)*(19/38) at the beginning, right ? But if we want to know the probability of picking an odd number before an even number at any time during the draw, it depends on what we've picked before. Let's say event A is picking an even number and event B is picking an odd number. Would the probability be P(A|B)=P(A∩B)/P(B) ?

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    Re: Probability to draw an odd number before an even number

    Does the question specify how many numbers to choose from the jar?
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: Probability to draw an odd number before an even number

    Quote Originally Posted by Buckeye View Post
    Does the question specify how many numbers to choose from the jar?
    All the numbers.

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    Re: Probability to draw an odd number before an even number


    I can't be sure. But I'm interpreting the problem to be the probability of drawing an odd number before the first even number. But this is probably incorrect. I want to reduce the problem to cards numbered 1,2,3,4,5 to illustrate. Aren't we considering the cases: OOOEE, OOE_ _, OE_ _ _ and then adding these probabilities together?
    Last edited by Buckeye; 02-10-2017 at 08:13 PM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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