Let us start with another problem:
I have $8$ kids which are $4$ pairs of brothers sitting randomly across $4$ tables, each table has two seats. What is the probability that no couple sits together?
Number the couples $1,2,3,4$ and let $B_{i}$ denote the event
that couple $i$ sits together.
Then to be found is $P\left(B_{1}^{\complement}\cap B_{2}^{\complement}\cap B_{3}^{\complement}\cap B_{4}^{\complement}\right)=1-P\left(B_{1}\cup B_{2}\cup B_{3}\cup B_{3}\right)$
Applying the principle of inclusion/exclusion and also symmetry we
find that this probability equals:
$$1-4P\left(B_{1}\right)+6P\left(B_{1}\cap B_{2}\right)-4P\left(B_{1}\cap B_{2}\cap B_{3}\right)+P\left(B_{1}\cap B_{2}\cap B_{3}\cap B_{4}\right)$$
Here we find:
- $P\left(B_{1}\right)=\frac{1}{7}$
- $P\left(B_{1}\cap B_{2}\right)=P\left(B_{2}\mid B_{1}\right)P\left(B_{2}\right)=\frac{1}{5}\frac{1}{7}$
- $P\left(B_{1}\cap B_{2}\cap B_{3}\right)=P\left(B_{3}\mid B_{1}\cap B_{2}\right)P\left(B_{1}\cap B_{2}\right)=\frac{1}{3}\frac{1}{5}\frac{1}{7}$
- $P\left(B_{1}\cap B_{2}\cap B_{3}\cap B_{4}\right)=P\left(B_{4}\mid B_{1}\cap B_{2}\cap B_{3}\right)P\left(B_{1}\cap B_{2}\cap B_{3}\right)=\frac{1}{1}\frac{1}{3}\frac{1}{5}\frac{1}{7}$
So we find: $$P\left(B_{1}^{\complement}\cap B_{2}^{\complement}\cap B_{3}^{\complement}\cap B_{4}^{\complement}\right)=1-4\cdot\frac{1}{7}+6\cdot\frac{1}{5}\frac{1}{7}-4\cdot\frac{1}{3}\frac{1}{5}\frac{1}{7}+\frac{1}{1}\frac{1}{3}\frac{1}{5}\frac{1}{7}=\frac{4}{7}$$
Now we step to the original problem.
Again number the couples $1,2,3,4,5$ and let $E_{i}$ denote the event that couple $i$ will sit together.
Further let $E$ denote the event that exactly one couple sits together.
Then $E$ is the union of the events:
- $E_{1}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}^{\complement}$
- $E_{1}^{\complement}\cap E_{2}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}^{\complement}$
- $E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}\cap E_{4}^{\complement}\cap E_{5}^{\complement}$
- $E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}\cap E_{5}^{\complement}$
- $E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}$.
These events are mutually excusive and equiprobable so that:
$$P\left(E\right)=5P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}\right)=5P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\mid E_{5}\right)P\left(E_{5}\right)$$
Here $P\left(E_{5}\right)=\frac{1}{9}$ because - after placing one
brother on a chair - there are $9$ chairs left and only $1$ of them
results in couple $1$ at the same table by placing the other brother.
Working under condition $E_{5}$ there are $4$ tables left for the
remaining $4$ couples and we are back in the problem that was solved
first.
So $P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\mid E_{5}\right)$ equals the $P\left(B_{1}^{\complement}\cap B_{2}^{\complement}\cap B_{3}^{\complement}\cap B_{4}^{\complement}\right)$ that was calculated there and we end up with:
$$P\left(E\right)=P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\mid E_{5}\right)P\left(E_{5}\right)=\frac{4}{7}\frac{1}{9}=\frac{4}{63}$$
Best Answer
Let's take as our sample space the number of ways of placing two people at each table, which is $$\binom{10}{2, 2, 2, 2, 2} = \binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit at each of those tables. That leaves four people and two tables. Choose which of those two tables the youngest of those four people sits at. One of the other three people will be his brother. To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining. The final two people must sit together at the remaining table. Hence, the number of favorable cases is $$\binom{5}{3} \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 2$$ Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is $$\frac{\dbinom{5}{3} \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 2}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}}$$