Csdrhrt

I have 10 kids which are 5 pairs of brothers sitting randomly across 5 tables, each table has two seats. What is the probability of only one couple sitting together?

Solving it with combinatorics alone (as it was done in this answer) is complicated because After pairing the first couple I have too many options. So I was trying to think of the complement event which is only $ 1 $ couple of brothers not sitting together, which is zero (because it can not be that only 1 couple are not sitting together). So what is the best way to solve it?

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 $Pleft(B_{1}^{complement}cap B_{2}^{complement}cap B_{3}^{complement}cap B_{4}^{complement}right)=1-Pleft(B_{1}cup B_{2}cup B_{3}cup B_{3}right)$

Applying the principle of inclusion/exclusion and also symmetry we
find that this probability equals:

$$1-4Pleft(B_{1}right)+6Pleft(B_{1}cap B_{2}right)-4Pleft(B_{1}cap B_{2}cap B_{3}right)+Pleft(B_{1}cap B_{2}cap B_{3}cap B_{4}right)$$

Here we find:

• $Pleft(B_{1}right)=frac{1}{7}$


• $Pleft(B_{1}cap B_{2}right)=Pleft(B_{2}mid B_{1}right)Pleft(B_{2}right)=frac{1}{5}frac{1}{7}$


• $Pleft(B_{1}cap B_{2}cap B_{3}right)=Pleft(B_{3}mid B_{1}cap B_{2}right)Pleft(B_{1}cap B_{2}right)=frac{1}{3}frac{1}{5}frac{1}{7}$


• $Pleft(B_{1}cap B_{2}cap B_{3}cap B_{4}right)=Pleft(B_{4}mid B_{1}cap B_{2}cap B_{3}right)Pleft(B_{1}cap B_{2}cap B_{3}right)=frac{1}{1}frac{1}{3}frac{1}{5}frac{1}{7}$

So we find: $$Pleft(B_{1}^{complement}cap B_{2}^{complement}cap B_{3}^{complement}cap B_{4}^{complement}right)=1-4cdotfrac{1}{7}+6cdotfrac{1}{5}frac{1}{7}-4cdotfrac{1}{3}frac{1}{5}frac{1}{7}+frac{1}{1}frac{1}{3}frac{1}{5}frac{1}{7}=frac{4}{7}$$

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:

$$Pleft(Eright)=5Pleft(E_{1}^{complement}cap E_{2}^{complement}cap E_{3}^{complement}cap E_{4}^{complement}cap E_{5}right)=5Pleft(E_{1}^{complement}cap E_{2}^{complement}cap E_{3}^{complement}cap E_{4}^{complement}mid E_{5}right)Pleft(E_{5}right)$$

Here $Pleft(E_{5}right)=frac{1}{9}$ 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 $Pleft(E_{1}^{complement}cap E_{2}^{complement}cap E_{3}^{complement}cap E_{4}^{complement}mid E_{5}right)$ equals the $Pleft(B_{1}^{complement}cap B_{2}^{complement}cap B_{3}^{complement}cap B_{4}^{complement}right)$ that was calculated there and we end up with:

$$Pleft(Eright)=Pleft(E_{1}^{complement}cap E_{2}^{complement}cap E_{3}^{complement}cap E_{4}^{complement}mid E_{5}right)Pleft(E_{5}right)=frac{4}{7}frac{1}{9}=frac{4}{63}$$