This doesn't directly answer your question, but provides an nice alternative method to solving this question
This might not exactly be the general method for solving a question like this given in a statistics class, but it's an possibly fun way to solve the question (something people do in math competitions)
Calculation through recurrence relations
Let $C_i$ denote the number of ways $i$ people can be seated in $2i$ chairs such that no two are adjacent.
Obviously for this question, we want to derive $$\frac{C_4}{\binom{8}{4}}$$
For $C_1$, there are two ways to seat one person in two chairs.
For $C_2$, you can reason that there is a total three ways.
For $C_3$, notice that if the third person is fixed/seated at the very end as so
$$-...-P$$
Where $-$ represents an chair and $P$ is a person, we realize that the total number of ways for the other two people to be seated so that no two are adjacent is equal to $C_2$.
If the third person is fixed/seated second to the very end as so
$$-...P-$$ Then it can be observed that this determines the seats of all the other people (try proving this yourself), thus giving us one combination for that seating for the third person.
So in conclusion, we see that $C_3=C_2+1$
In fact we can easily generalize this to $C_n=C_{n-1}+1$ by taking a more general case if you like.
But back to the question, for $C_4$, we know it is equal to $C_3+1$, which in turn is equal to $C_2+1+1=5$.
Knowing that, we can conclude that the answer to your question would be $$\frac {5}{\binom{8}{4}} = \frac {5}{70} = \frac {1}{14}$$