Suppose that $15$ people are divided in a random manner into two groups of 10 and 5. What is probability that two particular people A and B would be in the same group?
My attempt:
$$P(A \cap B)= \binom{15}{10} \binom{15}{5} $$
Am I doing the right thing?
On
We can calculate this probability without using combinations as follows:
Call the group of 10 people $G_{10}$ and the group of five people $G_5$. The probability that $A$ and $B$ are in the same group is:
$P(A \in G_{10} \land B \in G_{10}) + P(A \in G_5 \land B \in G_15)$
$=P(A \in G_{10})P(B \in G_{10} | A \in G_{10}) + P(A \in G_5)P(B \in G_5|A \in G_5)$
$=\frac{10}{15}\frac{9}{14} + \frac{5}{15}\frac{4}{14}=\frac{90+20}{210}=\frac{11}{21}$
The number of ways of selecting a group of ten people and a group of five people from a group of $15$ people is $$\binom{15}{10}\binom{5}{5} = \binom{15}{10}$$ since once we choose ten people to be in one group, the other five must be in the other group.
If $A$ and $B$ are both in the group of ten, then eight of the other thirteen people must also be in that group. Thus, the number of ways that $A$ and $B$ could both be selected for the group of ten is $$\binom{2}{2}\binom{13}{8}$$
If $A$ and $B$ are both in the group of five, then three of the other thirteen people must also be in that group. Thus, the number of ways that $A$ and $B$ could both be selected for the group of five is $$\binom{2}{2}\binom{13}{3}$$
Since these events are mutually exclusive, the number of favorable cases is $$\binom{2}{2}\binom{13}{8} + \binom{2}{2}\binom{13}{3}$$
Hence, the probability that $A$ and $B$ are in the same group is $$\frac{\dbinom{2}{2}\dbinom{13}{8} + \dbinom{2}{2}\dbinom{13}{3}}{\dbinom{15}{10}}$$
Note: In the comments, you arrived at the answer $$\frac{\dbinom{2}{2}\dbinom{13}{8}}{\dbinom{15}{10}} \cdot \frac{\dbinom{2}{2}\dbinom{13}{3}}{\dbinom{15}{5}}$$
You multiplied when you should have added.
The Multiplication Principle states that if there are $m$ ways of performing one task and $n$ ways of performing another task that can be performed independently of the first task, then there are $mn$ ways of performing both tasks. We used the Multiplication Principle here when we calculated the number of ways of selecting both $A$ and $B$ to be in the group of five. We chose both $A$ and $B$ and three of the other thirteen people in $$\binom{2}{2}\binom{13}{3}$$ ways. However, it does not apply to the number of ways that both $A$ and $B$ could be selected to be in the group of $10$ or the group of $5$ since these events cannot happen simultaneously.
The word and is an indication that you should multiply; the word or is an indication that you should add.
The Addition Principle states that if one event can occur in $m$ ways and another event that cannot occur at the same time can occur in $n$ ways, then the number of ways that one of the events could occur is $m + n$. Since it is not possible for $A$ and $B$ to simultaneously be in the group of $10$ and the group of $5$, you should have added the probability that $A$ and $B$ are both in the group of $10$ and the probability that they are both in the group of $5$. Had you done so, you would have obtained $$\frac{\dbinom{2}{2}\dbinom{13}{8}}{\dbinom{15}{10}} + \frac{\dbinom{2}{2}\dbinom{13}{3}}{\dbinom{15}{5}}$$ This is equivalent to my answer since $$\binom{15}{10} = \frac{15!}{10!5!} = \frac{15!}{5!10!} = \binom{15}{5}$$
In general, $$\binom{n}{k} = \binom{n}{n - k}$$ since $$\binom{n}{n - k} = \frac{n!}{(n - k)![n - (n - k)]!} = \frac{n!}{(n - k)!k!} = \frac{n!}{k!(n - k)!} = \binom{n}{k}$$