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$10$ kids are grouped into an A team with $5$ kids and a B team with five kids. If the group consists of five girls and five boys, what is the probability that all girls will end up on the same team

Here is my thought process for this problem: Choosing first person for Team A, there is a $\frac{5}{10}$ chance girl will be chosen, choosing next person for Team B, $\frac49$ chance of being girl, etc. For simplicity I'll call this $R$.

I believe the solution is $2\cdot R$ since there are two possible teams and either could have the all girl composition

is this correct?

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    $\begingroup$ Yes. Note: $R = \frac{1}{\binom{10}{5}}$, which is an easier way to see/write the value you describe. Picking the five people for team $A$ simultaneously, you can either pick all five girls or pick all five boys. When order doesn't matter, that is a total of two options. The total number possible where order doesn't matter regardless of all girls on one team is $\binom{10}{5}$. Dividing takes care of the rest. $\endgroup$ – JMoravitz May 24 '16 at 22:59
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    $\begingroup$ Hint: I guarantee that one girl (call her "Amy") will be on one of the two teams. What is the probability that the other four girls will also be on her team? $\endgroup$ – Graham Kemp May 25 '16 at 0:04
  • $\begingroup$ Your method is correct, and the answer you get is $2 / {10 \choose 5} = 1/126$, which is also equal to $1/{9 \choose 4}$. $\endgroup$ – svsring May 25 '16 at 6:39
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Pick any one of the girls.   She will be on one of the two teams but who cares which.   What we are after is the probability that the other four girls, selected from the other nine players, will also be on her team.

$$\dfrac{1}{\binom{9}{4}}$$

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There are $\frac{10\cdot9\cdot8\cdot7\cdot6}{5\cdot4\cdot3\cdot2\cdot1} = 252$ ways to select team A. Team B is whoever is not on team A. Team A could be all all girls or team A could be all boys (making team B all girls)

$\frac{2}{252} = \frac{1}{126}$

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If $R=\frac{5}{10} + \frac{4}{9} + \frac{3}{8} + \frac{2}{7} + \frac{1}{6}$ then the answer is $2R$.

But this problem is set up to use a hypergeometric distribution. Let $Y=$( # of girls on team A ). In general for a hypergeometric distribtion $P[Y=k]=\dfrac{(_KC_k)(_{(N-K)}C_{(n-k)})}{_{N}C_n}$ . In this case $K$ represents the number of girls total (5 of them) and $N$ is the total amount of people (10 of them) and $k$ is the number of girls we want to select (5 of them) and $n$ is the total amount of people we want to select (5 of them). We want to find the probability that the five girls are on the same team. This is $P[Y=5]+P[Y=0]$ since the five girls can be on team A or team B . So $P[Y=5] + P[Y=0] = \frac{(_5C_5)(_5C_0)}{_{10}C_5} + \frac{(_5C_0)(_5C_5)}{_{10}C_5} = 2 \cdot(.00397) $

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  • $\begingroup$ $\mathsf P(Y=5)$ is the probability that all of the girls are on a specific team. $~$ The post requested $\mathsf P(Y=5 \cup Y=0)$, that being the probability that all of the girls are on one team or the other team. $~$ That is indeed twice that. $\endgroup$ – Graham Kemp May 25 '16 at 2:01
  • $\begingroup$ Thanks for clarifying that @grahamkemp $\endgroup$ – alpastor May 25 '16 at 3:25
  • $\begingroup$ You mean the product (rather than the sum) of those five ratios. If you take the sum, the answer you get is incorrect and in fact larger than 1. $\endgroup$ – svsring May 25 '16 at 6:38
  • $\begingroup$ $\sum_{i=1}^5P[Y=i]=1$ @svsring $\endgroup$ – alpastor May 25 '16 at 17:45
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I take it that if all the girls are on team $A \;or\;B,$ the stipulation is met.

The first girl can be placed anywhere, the next has $4$ favorable slots in $9$ remaining, and so on, thus

$Pr = \dfrac49\dfrac38\dfrac27\dfrac16$

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