A group of $n ≥ 4$ people, including Alice and Bob, are seated at a round table, at random (where all the possible ways are equally likely).
(a) What is the probability that Alice and Bob sit next to each other?
*(b) If Cindy also sits next to the table, what is the probability that Bob sits next to either Alice or Cindy (or both)?
For part a, I am pretty confident:
$$2\frac{(n-2)!}{(n-1)!} = \frac{2}{n-1}$$
But I am not sure am I right for part b, my solution:
$$4\frac{(n-3)!}{(n-1)!}$$
Thank you!
Best Answer
Let Bob sit down as first.
a)
The probability that Alice will choose for one of the chairs next to Bob is:$$\frac2{n-1}$$
This simply because $2$ of the open $n-1$ chairs are next to Bob. Likewise thinking we can solve:
b)
The probability that Alice will not choose for one of the $2$ chairs next to Bob is $\frac{n-3}{n-1}$.
If - after that - Cindy takes her seat then the probability that Cindy will not choose for one of the $2$ chairs next to Bob is $\frac{n-4}{n-2}$.
That gives chance: $$1-\frac{n-3}{n-1}\frac{n-4}{n-2}$$ that the described event will not occur, i.e. that Alice or Cindy (or both) will sit next to Bob.