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?

  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Dec 1 '18 at 1:23
  • $\begingroup$ You should learn how to properly format the post. $\endgroup$ – Ya G Dec 1 '18 at 1:55
  • $\begingroup$ @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. $\endgroup$ – Shaun Dec 1 '18 at 2:09
  • 1
    $\begingroup$ Thanks, @Shaun. $\endgroup$ – N. F. Taussig Dec 1 '18 at 10:32

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 ${15\choose 5}{10\choose 5}{5\choose 5}$.

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


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 $$P(A\cap B^c)=P(A)-P(A\cap B)$$. 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 $$P(A\cap B)=\frac{{12\choose2}}{{14\choose4}}\approx0.0659$$ Use the same principle, to find $P(A)$, (which is easier) and then you are done.


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