[Math] Probability that Ken and John set next to each other

Question

A group of ten people sits down, uniformly at random, around a table. Ken
and John are part of this group. Determine the probability that Ken and John sit
next to each other.

There are $10!$ ways to arrange the seating for everyone, there are 10 possible ways for John and Ken to sit together.

$$\operatorname{Pr}(J\ \&\ K ) = \frac{10}{10!} = \frac{1}{9!}$$

Am I correct?

Answer 1

Two of nine people sit next to John. The probability that Ken is one of these two is $\frac29$.