Probabilistic method: what is the probability of the event that randomly selected composition of $n$ Can anyone help me out with the following question?
Q. What is the probability of the event that randomly selected composition of $n$ has a second part and that second part is $1$?
I know that the expected number, $E(X)$ that the first part of a randomly selected composition of $n$ is $2- 1/2^{n-1}$.
This problem is from 'A walk through Combinatorics' by Bona.
Thanks,

Obviously for $n=1$ the probability is $0$. I suggest that you do some experimenting with small values of $n$, say $n=2,3,4$, and $5$. The compositions that meet the requirement are as follows.
$n=2$: $$1+1$$
$n=3$: $$\begin{align*}&1+1+1\\&2+1\end{align*}$$
$n=4$: $$\begin{align*}&1+1+1+1\\&1+1+2\\&2+1+1\\&3+1\end{align*}$$
$n=5$:
$$\begin{align*}
&1+1+1+1+1\\
&1+1+1+2\\
&1+1+2+1\\
&2+1+1+1\\
&1+1+3\\
&3+1+1\\
&2+1+2\\
&4+1
\end{align*}$$
Presumably you know already that there are $2^{n-1}$ compositions of $n$, so with these experimental data you ought to be able to make a good conjecture. And once you’ve made that conjecture, ask yourself why the number of ‘good’ compositions of $n$ is the same as ...
