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Given the following question:

$12$ people are randomly divided into $3$ groups $g_1, g_2, g_3$.

$g_1$ has exactly $3$ members , $g_2$ has exactly $4$ members, $g_3$ has exactly $5$ members.

Each person $p_n$ belongs to only one group.

Let $p_i$, $p_j$ be persons , what is the probability that they're both in the same group?

Can you please explain how to approach such question? Thanks!

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In this problem it's easier to first view it as three possible situations

$\frac{3}{12}\times\frac{2}{11}$ Chance that they are both placed in first group

$\frac{4}{12}\times\frac{3}{11}$ Chance that they are both placed in second group

$\frac{5}{12}\times\frac{4}{11}$ Chance that they are both placed in third group

The probability that they are in same group is the sum of these three possibilities. $\frac{6}{132} + \frac{12}{132} + \frac{20}{132} = \frac{38}{132} = \frac{19}{66}$

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  • $\begingroup$ Why probability both are in $g_1$ is : $\frac{3}{12} \cdot \frac{2}{11}$? $\endgroup$
    – Billie
    Nov 1 '14 at 1:29
  • $\begingroup$ There are 12 possible places for Person A to go, 3 of them in group 1, hence (3/12). There are then 11 places for the Person B to go (Person A has already chosen a spot), 2 of them in group 1(Person A is already occupying one of the three places there), hence (2/11). $\endgroup$
    – TimD1
    Nov 2 '14 at 1:22
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The number of arrangements altogether is $$\frac{12!}{3!\,4!\,5!}\ .$$ The number of arrangements in which both $p_i,p_j$ are in group 1 is $$\frac{10!}{1!\,4!\,5!}\ .$$ So the probability that both these people are in group 1 is $$\frac{10!}{1!\,4!\,5!} \frac{3!\,4!\,5!}{12!}\ .$$ See if you can simplify this and then cover the other possible cases too.

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This seems like a straightforward application of addition principle- add the probabilities of them both being in $g_1, g_2, g_3$.

Probability both are in $g_1$ is: $\frac{3}{12} \cdot \frac{2}{11}=\frac{1}{22}$. Applying similar logic, find the probabilities of when both are in $g_2, g_3$, and add all these probabilities together.

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  • $\begingroup$ Why probability both are in $g_1$ is : $\frac{3}{12} \cdot \frac{2}{11}$? $\endgroup$
    – Billie
    Nov 1 '14 at 1:26
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First we need to count all the possible ways they can arranged. Let's assume that the order of people in a same group does not matter. Now, the number of ways they can be arranged are 12 choose 3 times 9 choose 4 and the rest is determined.

Now, we count all the ways they can be put in a group.

Case 1: they are put in $g_1$.

We are left to choose 1 from the 10 people to form $g_1$. And 9 choose 4 for the second group and the last group is determined.

Case 2:they are put in $g_2$.

Similarly, we have 10 choose 2 times 8 choose 3 times and $g_3$ is determined.

Case 3: they are put in $g_3$.

I bet you already know how to calculate this. 10 choose 3 times 7 choose 3 and $g_2$ is determined.

Adding all the cases up, and divide by the number of all possible cases. We obtain the probability.

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  • $\begingroup$ This is a combinatorics approach... a bit messier $\endgroup$
    – Kun
    Nov 1 '14 at 1:13

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