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Math Help - Another probability!

  1. #1
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    Post Another probability!

    Each of the numbers 1-8 is written on a piece of paper and placed in a hat. If two numbers are selected at random, find the probability that both numbers selected are even. (The papers are not replaced.)
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  2. #2
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    Re: Another probability!

    Well hi there!

    Do you know the probability of selecting an even number out of 1 to 8? Obviously it is

    p(even number out of 1-8) = ...

    Now imagine you already selected an even number. The papers are not replaced, so there are 7 papers left. There are still 4 of them labeled with an uneven number. In addition there are 3 even numbers left, because you have already selected an even number in the first draw. So it is

    p(even number out of 7 remaining papers) = 3/7.

    p(both selected numbers are even) = p(even number out of 1-8) * p(even number out of 7 remaining papers)

    Do you know what p(even number out of 1-8) is equal to?
    Thanks from alexgintom
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  3. #3
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    Re: Another probability!

    Hello, alexgintom!

    I assume you area familiar with Combinations.;


    Each of the numbers 1-8 is written on a piece of paper and placed in a hat.
    If two numbers are selected at random without replacement,
    . . find the probability that both numbers selected are even.

    There are 8 numbers: 4 even and 4 odd.

    There are: . _8C_2 \,=\,28\text{ ways to draw 2 numbers.}

    There are: . _4C_2 \,=\,6\text{ ways to draw 2 even numbers.}

    Therefore: . P(\text{both even}) \:=\:\frac{6}{28} \:=\:\frac{3}{14}
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