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Math Help - A jar contains 7 balls numbered from 1 to 7

  1. #1
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    A jar contains 7 balls numbered from 1 to 7

    A jar contains 7 balls numbered from 1 to 7. If a ball is selected at random, and the number is noted down, and a second ball is selected without replacing the first, what is the probability that the two-digit number is even?

    My Solution: if the number has to be even then its last digit is either 2,4,6.
    Odd Set = {1,3,5,7}
    Even Set = {2,4,6}
    There are two possibilities to have our desired outcome:

    I) Either both balls selected are even numbers therefore (C(1,3)*C(1,2)) / 7*6 = 6/42

    II) The first one is odd but the other is even therefore (C(1,4)(1,3)) / 7*6 = 12/42

    The result is : 6/42 + 12/42 = 18/42

    Is this correct? Is there any easy way? Thank you
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  2. #2
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    I would think you would just do:

    \displaystyle \frac{4}{7}*\frac{1}{2}=\frac{4}{14}+\frac{3}{7}*\  frac{1}{3}=\frac{2}{7}+\frac{1}{7}=\frac{3}{7}

    The end results are the same but the odd results differ.
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  3. #3
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    I'm sorry but I don't understand. Would you please explain so I can understand the concept?
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  4. #4
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    Never mind. I mixed up what you said was odd and even sorry.
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