Question regarding probability.












-1














Old McDonald keeps 15 goats on his farm, among them a goat called Edna. One of the other goats is Edna's best friend.
Unfortunately, another of them is her worst enemy. One day, Old McDonald decides to exercise the goats in three separate
enclosures. Five randomly selected goats will be assigned to each enclosure. What is the probability that Edna will be assigned to
the same group as her best friend, but not her worst enemy?



I tried doing 1*4/14*10/13. Is this the correct way to do this?






probability







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  • Please edit the question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Dec 1 at 1:23










  • You should learn how to properly format the post.
    – Ya G
    Dec 1 at 1:55










  • @N.F.Taussig If you type [edit], it links to the question at hand as if one has clicked on the automatic "edit" link. Look: edit.
    – Shaun
    Dec 1 at 2:09






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    Thanks, @Shaun.
    – N. F. Taussig
    Dec 1 at 10:32
















-1














Old McDonald keeps 15 goats on his farm, among them a goat called Edna. One of the other goats is Edna's best friend.
Unfortunately, another of them is her worst enemy. One day, Old McDonald decides to exercise the goats in three separate
enclosures. Five randomly selected goats will be assigned to each enclosure. What is the probability that Edna will be assigned to
the same group as her best friend, but not her worst enemy?



I tried doing 1*4/14*10/13. Is this the correct way to do this?






probability







share|cite|improve this question
























  • Please edit the question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Dec 1 at 1:23










  • You should learn how to properly format the post.
    – Ya G
    Dec 1 at 1:55










  • @N.F.Taussig If you type [edit], it links to the question at hand as if one has clicked on the automatic "edit" link. Look: edit.
    – Shaun
    Dec 1 at 2:09






  • 1




    Thanks, @Shaun.
    – N. F. Taussig
    Dec 1 at 10:32














-1












-1








-1







Old McDonald keeps 15 goats on his farm, among them a goat called Edna. One of the other goats is Edna's best friend.
Unfortunately, another of them is her worst enemy. One day, Old McDonald decides to exercise the goats in three separate
enclosures. Five randomly selected goats will be assigned to each enclosure. What is the probability that Edna will be assigned to
the same group as her best friend, but not her worst enemy?



I tried doing 1*4/14*10/13. Is this the correct way to do this?






probability







share|cite|improve this question















Old McDonald keeps 15 goats on his farm, among them a goat called Edna. One of the other goats is Edna's best friend.
Unfortunately, another of them is her worst enemy. One day, Old McDonald decides to exercise the goats in three separate
enclosures. Five randomly selected goats will be assigned to each enclosure. What is the probability that Edna will be assigned to
the same group as her best friend, but not her worst enemy?



I tried doing 1*4/14*10/13. Is this the correct way to do this?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 at 1:27

























asked Dec 1 at 1:16









Jmaxmanblue

18017




18017












  • Please edit the question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Dec 1 at 1:23










  • You should learn how to properly format the post.
    – Ya G
    Dec 1 at 1:55










  • @N.F.Taussig If you type [edit], it links to the question at hand as if one has clicked on the automatic "edit" link. Look: edit.
    – Shaun
    Dec 1 at 2:09






  • 1




    Thanks, @Shaun.
    – N. F. Taussig
    Dec 1 at 10:32


















  • Please edit the question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Dec 1 at 1:23










  • You should learn how to properly format the post.
    – Ya G
    Dec 1 at 1:55










  • @N.F.Taussig If you type [edit], it links to the question at hand as if one has clicked on the automatic "edit" link. Look: edit.
    – Shaun
    Dec 1 at 2:09






  • 1




    Thanks, @Shaun.
    – N. F. Taussig
    Dec 1 at 10:32
















Please edit the question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Dec 1 at 1:23




Please edit the question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Dec 1 at 1:23












You should learn how to properly format the post.
– Ya G
Dec 1 at 1:55




You should learn how to properly format the post.
– Ya G
Dec 1 at 1:55












@N.F.Taussig If you type [edit], it links to the question at hand as if one has clicked on the automatic "edit" link. Look: edit.
– Shaun
Dec 1 at 2:09




@N.F.Taussig If you type [edit], it links to the question at hand as if one has clicked on the automatic "edit" link. Look: edit.
– Shaun
Dec 1 at 2:09




1




1




Thanks, @Shaun.
– N. F. Taussig
Dec 1 at 10:32




Thanks, @Shaun.
– N. F. Taussig
Dec 1 at 10:32










2 Answers
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1














Your work (if indeed) is correct. Let Edna be in any group. The probability of selecting her best friend is $frac1{14}$ and there are $4$ places in Edna's group to position the friend. The probability of selecting her worst enemy is $frac1{13}$ and there are $10$ places in other two groups to position the enemy. Hence, the result.



Alternatively, the total number of allocations is ${15choose 5}{10choose 5}{5choose 5}$.



There are ${3choose 1}$ ways to place Edna and her best friend together in the same group (call it Friendship group). There are ${2choose 1}$ ways to place her worst enemy in other two groups (call it Enemy group). There are ${12choose 3}$ ways to select members for the Friendship group. There are ${9choose 4}$ ways to select for the Enemy group. There are ${5choose 5}$ ways to select for the Neutral group. In summary:
$$P(text{Edna with best friend and without worst enemy})=frac{{3choose 1}{2choose 1}{12choose 3}{9choose 4}{5choose 5}}{{15choose 5}{10choose 5}{5choose 5}}=frac{20}{91}.$$






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    0














    This looks like a homework problem you just don't want to take time solve. I will give you an idea.



    Let $A$ be an event where Edna is grouped with her best friend, and let $B$ be an event where she is with her enemy. Then, We want $$P(Acap B^c)=P(A)-P(Acap B)$$.
    Suppose Edna is already in a group. Then, there are 4 spots left. Then, out of remaining 14 goats, the number of ways to choose 4 goats is $14 choose 4$. Also, since the group that contains Edna, best friend, and enemy is determined by which two of the remaining 12 goats are in the group, we have to use $12 choose 2$. Thus, we have $$P(Acap B)=frac{{12choose2}}{{14choose4}}approx0.0659$$ Use the same principle, to find $P(A)$, (which is easier) and then you are done.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      active

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      1














      Your work (if indeed) is correct. Let Edna be in any group. The probability of selecting her best friend is $frac1{14}$ and there are $4$ places in Edna's group to position the friend. The probability of selecting her worst enemy is $frac1{13}$ and there are $10$ places in other two groups to position the enemy. Hence, the result.



      Alternatively, the total number of allocations is ${15choose 5}{10choose 5}{5choose 5}$.



      There are ${3choose 1}$ ways to place Edna and her best friend together in the same group (call it Friendship group). There are ${2choose 1}$ ways to place her worst enemy in other two groups (call it Enemy group). There are ${12choose 3}$ ways to select members for the Friendship group. There are ${9choose 4}$ ways to select for the Enemy group. There are ${5choose 5}$ ways to select for the Neutral group. In summary:
      $$P(text{Edna with best friend and without worst enemy})=frac{{3choose 1}{2choose 1}{12choose 3}{9choose 4}{5choose 5}}{{15choose 5}{10choose 5}{5choose 5}}=frac{20}{91}.$$






      share|cite|improve this answer


























        1














        Your work (if indeed) is correct. Let Edna be in any group. The probability of selecting her best friend is $frac1{14}$ and there are $4$ places in Edna's group to position the friend. The probability of selecting her worst enemy is $frac1{13}$ and there are $10$ places in other two groups to position the enemy. Hence, the result.



        Alternatively, the total number of allocations is ${15choose 5}{10choose 5}{5choose 5}$.



        There are ${3choose 1}$ ways to place Edna and her best friend together in the same group (call it Friendship group). There are ${2choose 1}$ ways to place her worst enemy in other two groups (call it Enemy group). There are ${12choose 3}$ ways to select members for the Friendship group. There are ${9choose 4}$ ways to select for the Enemy group. There are ${5choose 5}$ ways to select for the Neutral group. In summary:
        $$P(text{Edna with best friend and without worst enemy})=frac{{3choose 1}{2choose 1}{12choose 3}{9choose 4}{5choose 5}}{{15choose 5}{10choose 5}{5choose 5}}=frac{20}{91}.$$






        share|cite|improve this answer
























          1












          1








          1






          Your work (if indeed) is correct. Let Edna be in any group. The probability of selecting her best friend is $frac1{14}$ and there are $4$ places in Edna's group to position the friend. The probability of selecting her worst enemy is $frac1{13}$ and there are $10$ places in other two groups to position the enemy. Hence, the result.



          Alternatively, the total number of allocations is ${15choose 5}{10choose 5}{5choose 5}$.



          There are ${3choose 1}$ ways to place Edna and her best friend together in the same group (call it Friendship group). There are ${2choose 1}$ ways to place her worst enemy in other two groups (call it Enemy group). There are ${12choose 3}$ ways to select members for the Friendship group. There are ${9choose 4}$ ways to select for the Enemy group. There are ${5choose 5}$ ways to select for the Neutral group. In summary:
          $$P(text{Edna with best friend and without worst enemy})=frac{{3choose 1}{2choose 1}{12choose 3}{9choose 4}{5choose 5}}{{15choose 5}{10choose 5}{5choose 5}}=frac{20}{91}.$$






          share|cite|improve this answer












          Your work (if indeed) is correct. Let Edna be in any group. The probability of selecting her best friend is $frac1{14}$ and there are $4$ places in Edna's group to position the friend. The probability of selecting her worst enemy is $frac1{13}$ and there are $10$ places in other two groups to position the enemy. Hence, the result.



          Alternatively, the total number of allocations is ${15choose 5}{10choose 5}{5choose 5}$.



          There are ${3choose 1}$ ways to place Edna and her best friend together in the same group (call it Friendship group). There are ${2choose 1}$ ways to place her worst enemy in other two groups (call it Enemy group). There are ${12choose 3}$ ways to select members for the Friendship group. There are ${9choose 4}$ ways to select for the Enemy group. There are ${5choose 5}$ ways to select for the Neutral group. In summary:
          $$P(text{Edna with best friend and without worst enemy})=frac{{3choose 1}{2choose 1}{12choose 3}{9choose 4}{5choose 5}}{{15choose 5}{10choose 5}{5choose 5}}=frac{20}{91}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 at 5:23









          farruhota

          19k2736




          19k2736























              0














              This looks like a homework problem you just don't want to take time solve. I will give you an idea.



              Let $A$ be an event where Edna is grouped with her best friend, and let $B$ be an event where she is with her enemy. Then, We want $$P(Acap B^c)=P(A)-P(Acap B)$$.
              Suppose Edna is already in a group. Then, there are 4 spots left. Then, out of remaining 14 goats, the number of ways to choose 4 goats is $14 choose 4$. Also, since the group that contains Edna, best friend, and enemy is determined by which two of the remaining 12 goats are in the group, we have to use $12 choose 2$. Thus, we have $$P(Acap B)=frac{{12choose2}}{{14choose4}}approx0.0659$$ Use the same principle, to find $P(A)$, (which is easier) and then you are done.






              share|cite|improve this answer


























                0














                This looks like a homework problem you just don't want to take time solve. I will give you an idea.



                Let $A$ be an event where Edna is grouped with her best friend, and let $B$ be an event where she is with her enemy. Then, We want $$P(Acap B^c)=P(A)-P(Acap B)$$.
                Suppose Edna is already in a group. Then, there are 4 spots left. Then, out of remaining 14 goats, the number of ways to choose 4 goats is $14 choose 4$. Also, since the group that contains Edna, best friend, and enemy is determined by which two of the remaining 12 goats are in the group, we have to use $12 choose 2$. Thus, we have $$P(Acap B)=frac{{12choose2}}{{14choose4}}approx0.0659$$ Use the same principle, to find $P(A)$, (which is easier) and then you are done.






                share|cite|improve this answer
























                  0












                  0








                  0






                  This looks like a homework problem you just don't want to take time solve. I will give you an idea.



                  Let $A$ be an event where Edna is grouped with her best friend, and let $B$ be an event where she is with her enemy. Then, We want $$P(Acap B^c)=P(A)-P(Acap B)$$.
                  Suppose Edna is already in a group. Then, there are 4 spots left. Then, out of remaining 14 goats, the number of ways to choose 4 goats is $14 choose 4$. Also, since the group that contains Edna, best friend, and enemy is determined by which two of the remaining 12 goats are in the group, we have to use $12 choose 2$. Thus, we have $$P(Acap B)=frac{{12choose2}}{{14choose4}}approx0.0659$$ Use the same principle, to find $P(A)$, (which is easier) and then you are done.






                  share|cite|improve this answer












                  This looks like a homework problem you just don't want to take time solve. I will give you an idea.



                  Let $A$ be an event where Edna is grouped with her best friend, and let $B$ be an event where she is with her enemy. Then, We want $$P(Acap B^c)=P(A)-P(Acap B)$$.
                  Suppose Edna is already in a group. Then, there are 4 spots left. Then, out of remaining 14 goats, the number of ways to choose 4 goats is $14 choose 4$. Also, since the group that contains Edna, best friend, and enemy is determined by which two of the remaining 12 goats are in the group, we have to use $12 choose 2$. Thus, we have $$P(Acap B)=frac{{12choose2}}{{14choose4}}approx0.0659$$ Use the same principle, to find $P(A)$, (which is easier) and then you are done.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 at 1:51









                  Ya G

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