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I have seen a problem in Probability that I am stuck : Imagine that we have $12$ students in class and among this students there are $3$ honor students. Say that a teacher wants to assign a group project and wants to balance the groups out by forming 3 groups of students with exactly one honor student.

  • What is the probability of forming groups of 4 students with exactly one honor student in each group, if the students are selected randomly?

P.S. I have thought that the number of different ways to select groups of $4$ students are $$\binom{12}{4}\binom{8}{4}\binom{4}{4}$$ And I reasoned that the number of ways of selecting groups with exactly one honor student could be $$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$ The answer indicates approximately $.60$ as the answer. What am I doing wrong? Thank you.

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  • $\begingroup$ I have added a Remark that carries out your analysis, with small modification. $\endgroup$ – André Nicolas Nov 14 '13 at 22:03
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Imagine that we line up the students at random, and assign the first $4$ to one group, the next $4$ to another, and the last $4$ to another.

We can analyze as follows. We have $3$ copies of the letter $H$, and $9$ copies of the letter $M$. We make a $12$-letter word. What is the probability there will be an $H$ among the first $4$ letters, and an $H$ among the next $4$, and an $H$ among the last $4$?

There are $\binom{12}{3}$ equally likely ways to place the $3$ $H$'s.

The number of choices with one $H$ in the first $4$, another $H$ in the next $4$, another in the next $4$ is $4^3$.

Our required probability is therefore $\dfrac{4^3}{\binom{12}{3}}$.

Remark: Your number $\binom{12}{8}\binom{8}{4}\binom{4}{4}$ counts the number of ways to divide our $12$ people into three uniformed teams, the Blues, the Whites, and the Reds. That's fine.

Now we count the number of ways to divide into uniformed teams with an Honours student on each team. The Honours student of the Blues can be chosen in $\binom{3}{1}$ ways. The rest of the team can be chosen in $\binom{9}{3}$ ways. For every such choice, the Honours student in the Whites can be chosen in $\binom{2}{1}$ ways, and the rest of her team in $\binom{6}{3}$ ways. Now we are finished, although we can tack on a decorative $\binom{1}{1}\binom{3}{3}$ for a total of $\binom{3}{1}\binom{9}{3}\binom{2}{1}\binom{6}{3}\binom{1}{1}\binom{3}{3}$.

Now for the probability, divide.

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  • $\begingroup$ I really like your simple solution @André Nicolas. Thank you for your input... Now I know the answer key is wrong. $\endgroup$ – Heber Nov 14 '13 at 22:37
  • $\begingroup$ You are welcome. Your analysis was fine, apart from a little problem. $\endgroup$ – André Nicolas Nov 14 '13 at 23:20
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I just attempted to solve it and got $.2909$

My approach was select a group of $4$ randomly. There is a $.50909$ probability this group has exactly one honor student. If you get a group with exactly one honor student then select a group of $4$ randomly from the remaining $8$ students. The chance of this group containing $1$ honor student is $.57143$. If both of those situations obtain, a $.29091$ probability then the last group also contains one honor student.

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Herber - you were almost right, although I'm not sure where your 60% came from. Divide your bottom group by your top group - 1680/34650. The only thing you forgot was that after you've calculated how many groups can have 1 honor student, you still have to multiply by the possible ways of arranging those honor students, which is 3!=6.

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  • $\begingroup$ Thank you @Mitchell Kaplan . I really appreciate your feedback. $\endgroup$ – Heber Nov 14 '13 at 22:01

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