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4 girls and 8 boys are randomly divided into 3 groups of equal size What is the probability that there will be at least one girl in each group?

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    $\begingroup$ So, how far did you get when you tried to solve the problem? Where did you get stuck? $\endgroup$ – Gerry Myerson Apr 12 '15 at 13:20
  • $\begingroup$ Any thoughts about the answers that have been posted? $\endgroup$ – Gerry Myerson Apr 15 '15 at 5:59
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You can calculate the probability that all girls are in a only group, the probability that four girls are disposed in two groups and then you can use the inverse probability. The probability that all girls are in only group is: $$1\cdot \frac {3}{11}\cdot \frac{2}{10}\cdot {1}{9}$$ while the probability that all girls are disposed in two groups is:$$1\cdot 1\cdot \frac{6}{10}\cdot \frac{5}{9}.$$ Then you can use the inverse probability $$1-(1\cdot \frac{3}{11}\cdot \frac{2}{10}\cdot \frac{1}{9})-(1\cdot 1\cdot \frac{6}{10}\cdot \frac{5}{9}).$$

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  • $\begingroup$ It looks like you meant $1 \cdot \frac{3}{11} \cdot \frac{2}{10} \cdot \color{red}{\frac{1}{9}}$. $\endgroup$ – N. F. Taussig Apr 12 '15 at 15:31
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1. How many possibe ways are there to put $12$ people into $3$ groups with $4$ members each?


The first group we choose to fill has $\binom{12}{4}$ different configurations. The scound group has only $\binom{8}{4}$ and the third group consists of the remaining $4$ people. Hence the total amount is equal to $$\binom{12}{4} \cdot \binom{8}{4}$$

2. Now we have to calculate how many of the above configurations have at least one group without a girl.


  • 2.1 Only one group is without girls. If one group is girl-free then it has to be filled with boys and therefore we have to calculate the number of ways where we put 4 boys and 4 girls into the 2 remaining groups. This amount we multiply by 3 for there are three groups which can be girl-free hence: $$3 \cdot \binom{8}{4}$$

  • 2.2 Two groups are girl-free. This means we put all girls into one group. We again have three groups we could fill with girls. $$ 3\cdot 1$$

3 There are $3 \cdot \binom{8}{4} + 3\cdot 1$ sortings where at least one group is girl-free and the total amount of sorting is $\binom{12}{4} \cdot \binom{8}{4}$.

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