Finding the PMF of number of heads resulting from second round of tosses of coins There are $n$ coins and each one shows heads with probability $p$, independently of each of the other coins. Each coin which shows head is tossed again. What is the probability mass function of the number of heads resulting from the second round of tosses?  
My attempt is:  
Define $Y$ to be the number of heads in the first round of tosses and $X$ to be the number of heads in the second round of tosses.  
Then we have
$$\mathbb{P}(X=x) = \sum_{y=0}^n \mathbb{P}(X=x|Y=y)\mathbb{P}(Y=y) \\ = \sum_{y=0}^n \binom{y}x p^x (1-p)^{y-x} \times \binom{n}y p^y (1-p)^{n-y}.
$$
After attempting to simplify this, I couldn't get the given solution of
$$p(x) = \binom{n}x p^{2x}(1-p^2)^{n-x}.
$$
 A: Instead of looking at this as two independent rounds consider a single process: the probability of having heads 2 times in a row, which is achieved by each coin with a probability of $p^2$. This still remains a binomial model but we have $p$ replaced by $p^2$.Thus:
$P(X=x)=\binom {n}{x}p^{2x} (1-p^{2})^{n-x}$, where X is the number of heads in the second round
A: Sum $y$ from $x$ to $n$ as $\mathsf P(X=x,Y=y)\mathbf 1_{x>y}=0$ (since it is impossible to toss count more after the second elimination than the first.)
Everything else is algebraic rearrangements and binomial identities.
$\begin{align}\mathsf P(X=x)
\tag 1 & = \sum_{y=x}^n \mathsf P(X=x\mid Y=y)\cdot\mathsf P(Y=y)
\\ \tag 2 &= \sum_{y=x}^n\binom yx p^x(1-p)^{y-x}\cdot\binom nyp^y(1-p)^{n-y}
\\ \tag 3 &= p^x(1-p)^{n-x}\cdot\sum_{y=x}^n \binom{n}{y}\binom{y}{y-x} p^{y}
\\ \tag 4 &= p^x(1-p)^{n-x}\cdot\sum_{y=x}^n \binom{n}{n-x}\binom{n-x}{y-x} p^{y}
\\ \tag 5 &= \binom{n}{x}p^{2x}(1-p)^{n-x}\cdot\sum_{y=x}^n \binom{n-x}{y-x} p^{y-x}
\\ \tag 6 &= \binom{n}{x}p^{2x}(1-p)^{n-x}\cdot\sum_{j=0}^{n-x} \binom{n-x}{j} p^{j}
\\ \tag 7 &= \binom{n}{x}p^{2x}(1-p)^{n-x}\cdot(1+p)^{n-x}
\\ \tag 8 &= \binom{n}{x}(p^2)^{x}(1-p^2)^{n-x}
\end{align}$
