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An ordinary die is has six faces numbered from 1 to 6. If an ordinary die is thrown four times, find the probability of obtaining two even numbers and two odd numbers.

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    $\begingroup$ Any thoughts? Since Even/odd are equal probability you can just think of this as $4$ coin tosses out of which you want two Heads. If all else fails, enumeration is easy. $\endgroup$ – lulu Feb 21 '16 at 13:34
  • $\begingroup$ $\binom42\cdot\left(\frac36\right)^{2}\cdot\left(1-\frac36\right)^{4-2}$ $\endgroup$ – barak manos Feb 21 '16 at 13:54
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There are two outcomes with equal probability, so you want to divide combination $\binom{4}{2}=6$ (these are the cases you want) to the number of all possible outcomes (even or odd) which is $2^4$. So 6/16 should be the answer.

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