Question
Let $n$ and $k$ be integers such that n is even, $n\ge2$ and $1\le k\le n$. You are having a party where $n$ students attended.
a) $k$ of these $n$ students are politically correct and, thus, refuse to say Merry Christmas. Instead, they say Happy Holidays.
b) $n - k$ of these $n$ students do not care about political correctness and, thus, they say Merry Christmas.
Consider a uniformly random permutation of these n students. The positions in this permutation are numbered as $1,2,…,n$.
Define the random variable $X$,
$X$ = the number of positions with $i$ with 1<=$i$<=$\frac{n}{2}$ such that both students at positions $i$ and $2i$ are politically correct.
What is the expected value $E(X)$ of the random variable $X$? (Use indicator variables)
Options:
a) $n$ $.$ $\frac{k(k-1)}{n(n-1)}$
b) $n$ $.$ $\frac{(k-1)(k-2)}{n(n-1)}$
c) $\frac{n}{2}$ $.$ $\frac{k(k-1)}{n(n-1)}$
d) $\frac{n}{2}$ $.$ $\frac{(k-1)(k-2)}{n(n-1)}$
answer is c).
Attempt:
Indicator Variable:
$X$ $= 1$ if $i$ with 1<=$i$<=$\frac{n}{2}$ such that both students at positions $i$ and $2i$ are politically correct.
$X=0$ for all other cases
We need $E(X)$ = $\sum_{k=0}^{n/2} k . p(k)$
We have $\frac{n}{2}$ positions? but I can’t seem to find $p(k)$
There’s so much information given in this question that Im confused on how to break it down beyond the basic initial expected value steps.