If an ordinary six-sided die is thrown four times, find the probability of obtaining two even numbers and two...












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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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    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.
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    – lulu
    Feb 21 '16 at 13:34










  • $begingroup$
    $binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
    $endgroup$
    – barak manos
    Feb 21 '16 at 13:54


















1












$begingroup$


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.






probability







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    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










  • $begingroup$
    $binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
    $endgroup$
    – barak manos
    Feb 21 '16 at 13:54
















1












1








1





$begingroup$


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.






probability







share|cite|improve this question











$endgroup$




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.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 21 '16 at 13:36









N. F. Taussig

44.3k93357




44.3k93357










asked Feb 21 '16 at 13:10









user316442user316442

61




61








  • 3




    $begingroup$
    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










  • $begingroup$
    $binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
    $endgroup$
    – barak manos
    Feb 21 '16 at 13:54
















  • 3




    $begingroup$
    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










  • $begingroup$
    $binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
    $endgroup$
    – barak manos
    Feb 21 '16 at 13:54










3




3




$begingroup$
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




$begingroup$
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












$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– barak manos
Feb 21 '16 at 13:54






$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– 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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    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 21 '16 at 13:51









        TonyK

        42.6k355134




        42.6k355134










        answered Feb 21 '16 at 13:43









        n0pn0p

        101




        101






























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