probability - Finding the distribution of a n tossed fair coin

I am trying to solve the problem:

Consider a sequence of n tosses of a fair coin. Let X denote the number of heads, and Y denote the
number of isolated heads, that come up. (A head is an “isolated” head if it is immediately preceded

and followed by a tail, except in: position $1$ where a head is only followed by a tail, and position n
where the head is only preceded by a tail.) Additionally, let $X_i = 1$ if the ith coin toss results in
heads, and $X_i$ = 0 otherwise, i = 1, . . . , n. Similarly, $Y_j = 1$ if the jth coin toss results in an isolated
head, and $Y_j = 0\ otherwise, j = 1, . . . , n$.

How to find the distribution of $Y |X = 2$.

Here I think $X \sim Bin(n,1/2) \ and\ Y \sim Bin(n,1/8)$. I couldn't figure out the solution of this problem. Any help would be highly appreciated.

Answer

You either have $0$ isolated heads or $2$ isolated heads.

There are $\begin{pmatrix} n \\ 2 \end{pmatrix}$ ways to place exactly $2$ heads among the coin tosses.

Of which $n-1$ of these choices gives you non-isolated heads..

Hence

$$Pr(Y=0|X=2)=\frac{n-1}{\begin{pmatrix} n \\ 2 \end{pmatrix}}=\frac{2}{n}$$

$$Pr(Y=2|X=2)=1-Pr(Y=0|X=2)=\frac{n-2}{n}$$