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We are partitioning a group of $30$ people in $5$ groups of $6$ persons each. We have $13$ women and $17$ men in those $30$ people and randomly drawing those people gave us a men-only group. What are the odds of getting a group of the same gender?

A generic way is preferred, of course. So, let $N$ be the number of people to partition in $m$ groups while $G[k]$ is the number of people per gender.

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  • 1
    $\begingroup$ (This is a real problem and five people of the teaching staff are now spending the afternoon discussing possible solutions.) $\endgroup$ – Debilski Sep 11 '14 at 12:09
  • $\begingroup$ Are the groups different? $\endgroup$ – Snufsan Sep 11 '14 at 12:59
  • $\begingroup$ This is Prob(exactly one group of 6 men OR 2 groups Or 3 groups) = Prob(exactly one group of 6 men) + Prob(exactly 2 groups of 6 men) + Prob(exactly 3 groups of 6 men). As 5 groups, Prob(exactly one group of 6 men) = 5*Prob(first group has exactly 6 men). There are 17*16*15*14*13*12 ways to put 6 men in group one and 30*29*28*27*26*25 ways to put 6 people in group 1. So Prob(exactly one group of 6 men) = 5*(17*16*15*14*13*12)/(30*29*28*27*26*25) which is approx 0.104. The other two possibilities will have much smaller probability than this so the probability would be a bit above 0.1. I think.. $\endgroup$ – Paul Sep 11 '14 at 13:09
  • $\begingroup$ The title of the question and the question stem does not seem to match. $\endgroup$ – Satish Ramanathan Sep 11 '14 at 13:21
  • $\begingroup$ @Paul: I think $5*(17*16*15*14*13*12)/(30*29*28*27*26*25)$ may be the expected number of groups of $6$ men rather than the probability of exactly one, but I suspect it is close to the correct answer. $\endgroup$ – Henry Sep 11 '14 at 13:45
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Use the Principle of Inclusion/Exclusion

The probability of getting a group with all men is:

$$\begin{align} \mathsf P(M) & = \frac{{5\choose 1}{17\choose 6,11}{24\choose 6,6,6,6}-{5\choose 2}{17\choose 6,6,5}{18\choose 6,6,6}}{30\choose 6,6,6,6,6} \\[1ex] & = \frac{{5\times 17!\times 24!\over 11!}-{10\times 17!\times 18!\over 5!}}{30!} \\[1ex] & =\frac{59024}{570285} \\[1ex] & \approx 0.103{\small 5\tiny\dotsc} \end{align}$$

The probability of getting a group with all women is:

$$\begin{align} \mathsf P(W) & = \frac{{5\choose 1}{13\choose 6,7}{24\choose 6,6,6,6}-{5\choose 2}{13\choose 6,6,1}{18\choose 6,6,6}}{30\choose 6,6,6,6,6} \\[1ex] & = \frac{{5\times 13!\times 24!\over 7!}-{10\times 13!\times 18!\over 1!}}{30!} \\[1ex] & = \frac{19226}{1330665} \\[1ex] & \approx 0.0144{\small 5\tiny\dotsc} \end{align}$$

The probability of getting a group with all men and a group with all women is:

$$\begin{align} \mathsf P(M\cap W) & = \frac{{5\choose 1,1,3}{17\choose 6,11}{13\choose 6,7}{18\choose 6,6,6}-{5\choose 2,1,2}{17\choose 6,6,5}{13\choose 6,7}{12\choose 6,6}-{5\choose 1,2,2}{17\choose 6,11}{13\choose 6,6,1}{12\choose 6,6}+{5\choose 2,2,1}{17\choose 6,6,5}{13\choose 6,6,1}{6\choose 6}}{30\choose 6,6,6,6,6} \\[2ex] & = \frac{\frac{{20}\times{17!}{13!}{18!}}{11!7!}-\frac{{30}\times{17!}{13!}{12!}}{5!7!}-\frac{{30}\times{17!}{13!}{12!}}{11!}+\frac{{30}\times{17!}{13!}{6!}}{5!}}{30!} \\[2ex] & = \frac{6806}{1330665} \\[2ex] & \approx 0.00511{\small 5\tiny\dotsc} \end{align}$$

so thus the probability of a group of all men or one of all women is: $$\mathsf P(M\cup W)= \mathsf P(M)+\mathsf P(W)-\mathsf P(M\cap W)$$

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As a check on these analytical results, we can simulate this in R:

f = function(){
  is.man = matrix(sample(30) <= 17, nrow=5, ncol=6)
  man.gp.counts = rowSums(is.man)
  contains.all.man.group = sum(man.gp.counts == 6) >= 1
  return(contains.all.man.group)
}

table(replicate(10^6, f())

This returns

 FALSE   TRUE 
896547 103453 

The function f randomly groups the individuals into six groups and takes their gender (the individuals are numbered 1 to 30, of whom the first 17 are men); then it determines the number of men in each group, and returns TRUE if there is a group of all men.

The table shows that the probability of getting an all-male group is about 10.3% (on one million simulations), which agrees with Graham Kemp's answer.

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My humble try:

Total Number of ways denoted by B $$=\frac{{30\choose6}{24\choose6}{18\choose6}{12\choose6}{6\choose6}}{5!}$$

Number of ways atleast one group is of same gender denoted by A

enter image description here

The probability that you have atleast one group of same gender $= \frac{A}{B}$

Thanks

Satish

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  • $\begingroup$ $0.289\%$, i.e. $0.00289$, looks too low. The expected number of single gender groups is $5 \times \dfrac{{17\choose 6}+{13 \choose 6}}{{30 \choose 6}} \approx 0.11866$ so the probability of at least one will be slightly below this. $\endgroup$ – Henry Sep 11 '14 at 13:40

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