Prove the probability $P(X>0) \ge 0.39$. 
There are $n$ children in class. Each two children can be friends with probability of $p$ (so they are not friends with probability of $1-p$). Let $X$ the random variable denotes the number of children with no friends at all. Prove that when $n=10$ and $p=0.25$: $P(X>0) \ge 0.39$.

I already showed that:  


*

*$E(x) = n(1-p)^{n-1}$.

*$V(x) = n(1-p)^{n-1} + n(n-1)(1-p)^{2n-3} - n^2(1-p)^{2n-2}$.

*When $n=10$ and $p=0.25$: $P(x>0) \le 0.76$.


Now,


*

*I want to utilize Chebyshev's inequality but not sure how.

*I understood that the following can work as well: $$\sum_{n=1}^{10} Pr(x_i = 1) - \sum_{i\ne j} Pr(x_i=1,x_j=1) = \sum_{n=1}^{10} (1-0.25)^9 - {n \choose 2} 0.75^80.75^80.75$$


Could you explain this solution?
 A: We can find an expression for $P(X>0)$ as below.
Let $X_i$ be an indicator random variable associated with child $i$ and it takes value $1$ if the child has at least one friend and $0$ if the child has no friend. Then, $X_i\sim Bern(p'),\ p'=1-(1-p)^{n-1}$. Then $$P(X>0)=P\left(\bigcup_{i=1}^n\{X_i=0\}\right)\\ =\sum_{i=1}^n P(X_i=0)-\sum_{i<j}P\left(\{X_=0\}\cap\{X_j=0\}\right)+\cdots$$ Now, for a selection of $k$ children indexed as $\pi(1),\cdots,\ \pi(k)$ $$P\left(\bigcap_{i=1}^k\{X_{\pi(i)}=0\}\right)=P(\mbox{none of $\pi(i),\ i=1,2,\cdots,\ k$ has a single friend})$$ Let $A_{ij}$ denotes the event that a pair $i,j$ are not friends. Then $$P\left(\bigcap_{i=1}^k\{X_{\pi(i)}=0\}\right)=P\left(\bigcap_{i=1}^k\bigcap_{j\in [n]\setminus\{\pi(i)\}}A_{\pi(i),j}\right)$$ where $[n]:=\{1,2,\cdots,\ n\}$. Now if the events $A_{ij}$ are assumed to be mutually independent, then the event  $\bigcap_{i=1}^k\bigcap_{j\in [n]\setminus\{\pi(i)\}}A_{\pi(i),j}$ is the intersection of $\sum_{j=1}^k (n-j)=nk-\dfrac{k(k+1)}{2}$ mutually independent events each with probability $q=1-p$. Then, $$P\left(\bigcap_{i=1}^k\{X_{\pi(i)}=0\}\right)=q^{nk-\frac{k(k+1)}{2}}$$ Then, we have $$P(X>0)=\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}q^{nk-\frac{k(k+1)}{2}}$$
