# $12$ people $p_1, .. , p_{12}$ divided into $3$ groups, what is the probability that $p_i$ and $p_j$ are in the same group?

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!

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. $\frac{6}{132} + \frac{12}{132} + \frac{20}{132} = \frac{38}{132} = \frac{19}{66}$

• Why probability both are in $g_1$ is : $\frac{3}{12} \cdot \frac{2}{11}$? Nov 1 '14 at 1:29
• 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). Nov 2 '14 at 1:22

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.

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.

• Why probability both are in $g_1$ is : $\frac{3}{12} \cdot \frac{2}{11}$? Nov 1 '14 at 1:26

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.

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