If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?

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If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?



I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?



Any help would be much appreciated!






probability permutations







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    up vote
    1
    down vote

    favorite












    If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?



    I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?



    Any help would be much appreciated!






    probability permutations







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?



      I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?



      Any help would be much appreciated!






      probability permutations







      share|cite|improve this question















      If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?



      I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?



      Any help would be much appreciated!






      probability permutations




      probability permutations






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 52 mins ago









      amWhy

      190k27221433




      190k27221433










      asked 1 hour ago









      chloe loughan

      3219




      3219




















          2 Answers
          2






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          up vote
          5
          down vote



          accepted










          Let's calculate the probability, then convert that to odds.



          On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $dfrac12$.



          The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $dfrac 1 2$ of obtaining the desired result. So, we have:



          $$P(E,E,O) = dfrac 1 2 cdot dfrac 1 2 cdot dfrac 1 2 = dfrac 1 8$$



          Now, the probability of that not happening is $$1-dfrac 1 8 = dfrac 7 8$$



          So, the odds are 7:1 against the desired outcome.






          share|cite|improve this answer



























            up vote
            2
            down vote













            I think you might be over complicating things.



            It has to be even on the first two, the probability of this is $frac12 times frac12 = frac14.$



            The probability of odd on the third roll is also $frac12$ so your final probability is $frac18.$






            share|cite|improve this answer




















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              5
              down vote



              accepted










              Let's calculate the probability, then convert that to odds.



              On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $dfrac12$.



              The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $dfrac 1 2$ of obtaining the desired result. So, we have:



              $$P(E,E,O) = dfrac 1 2 cdot dfrac 1 2 cdot dfrac 1 2 = dfrac 1 8$$



              Now, the probability of that not happening is $$1-dfrac 1 8 = dfrac 7 8$$



              So, the odds are 7:1 against the desired outcome.






              share|cite|improve this answer
























                up vote
                5
                down vote



                accepted










                Let's calculate the probability, then convert that to odds.



                On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $dfrac12$.



                The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $dfrac 1 2$ of obtaining the desired result. So, we have:



                $$P(E,E,O) = dfrac 1 2 cdot dfrac 1 2 cdot dfrac 1 2 = dfrac 1 8$$



                Now, the probability of that not happening is $$1-dfrac 1 8 = dfrac 7 8$$



                So, the odds are 7:1 against the desired outcome.






                share|cite|improve this answer






















                  up vote
                  5
                  down vote



                  accepted







                  up vote
                  5
                  down vote



                  accepted






                  Let's calculate the probability, then convert that to odds.



                  On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $dfrac12$.



                  The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $dfrac 1 2$ of obtaining the desired result. So, we have:



                  $$P(E,E,O) = dfrac 1 2 cdot dfrac 1 2 cdot dfrac 1 2 = dfrac 1 8$$



                  Now, the probability of that not happening is $$1-dfrac 1 8 = dfrac 7 8$$



                  So, the odds are 7:1 against the desired outcome.






                  share|cite|improve this answer












                  Let's calculate the probability, then convert that to odds.



                  On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $dfrac12$.



                  The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $dfrac 1 2$ of obtaining the desired result. So, we have:



                  $$P(E,E,O) = dfrac 1 2 cdot dfrac 1 2 cdot dfrac 1 2 = dfrac 1 8$$



                  Now, the probability of that not happening is $$1-dfrac 1 8 = dfrac 7 8$$



                  So, the odds are 7:1 against the desired outcome.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 50 mins ago









                  InterstellarProbe

                  3,041721




                  3,041721




















                      up vote
                      2
                      down vote













                      I think you might be over complicating things.



                      It has to be even on the first two, the probability of this is $frac12 times frac12 = frac14.$



                      The probability of odd on the third roll is also $frac12$ so your final probability is $frac18.$






                      share|cite|improve this answer
























                        up vote
                        2
                        down vote













                        I think you might be over complicating things.



                        It has to be even on the first two, the probability of this is $frac12 times frac12 = frac14.$



                        The probability of odd on the third roll is also $frac12$ so your final probability is $frac18.$






                        share|cite|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          I think you might be over complicating things.



                          It has to be even on the first two, the probability of this is $frac12 times frac12 = frac14.$



                          The probability of odd on the third roll is also $frac12$ so your final probability is $frac18.$






                          share|cite|improve this answer












                          I think you might be over complicating things.



                          It has to be even on the first two, the probability of this is $frac12 times frac12 = frac14.$



                          The probability of odd on the third roll is also $frac12$ so your final probability is $frac18.$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 51 mins ago









                          MRobinson

                          79115




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