Coins and probability Bob has $n$ coins, each of which falls heads with the probability $p$. In the first round Bob tosses all coins, in the second round Bob tosses only those coins which fell heads in the first round. Let $R_i$ the number of coins which fell heads in the round $i$.


*

*What is the distribution law for $R_2$?

*Find $Corr(R_1,R_2)$

*How does correlation behave when $p→0$ and $p→1$? Why?

 A: Each coin falls head in the second round if and only it fell head in the first round (hence was tossed again) and fell head in the second round as well. Thus each coin falls head in the second round with probability $p^2$, and the number $R_2$ of these has binomial $(n,p^2)$ distribution.
Since $R_1$ is binomial $(n,p)$, $\mathbb E(R_1)=np$, $\mathrm{var}(R_1)=np(1-p)$ and $\mathbb E(R_1^2)=\mathrm{var}(R_1)+\mathbb E(R_1)^2=np(1-p)+n^2p^2$. Since $R_2$ is binomial $(n,p^2)$,  $\mathbb E(R_2)=np^2$ and  $\mathrm{var}(R_2)=np^2(1-p^2)$. For every $k$, conditionally on $R_1=k$, $R_2$ is binomial $(k,p)$, hence $\mathbb E(R_2\mid R_1)=R_1p$, and
$$
\mathbb E(R_1R_2)=\mathbb E(R_1\mathbb E(R_2\mid R_1))=p\mathbb E(R_1^2).
$$
Thus,
$$
\text{Cov}(R_1,R_2)=\mathbb E(R_1R_2)-\mathbb E(R_1)\mathbb E(R_2)=np^2(1-p),
$$
and
$$
\text{Corr}(R_1,R_2)=\frac{\text{Cov}(R_1,R_2)}{\sqrt{\text{var}(R_1)\text{var}(R_2)}}=\sqrt{\frac{p}{1+p}}.
$$
A: $P(R_2=i)=\sum_{j=i}^{n}{C(n,j)}{C(j,i)}p^{j+i}(1-p)^{n-i}$
Is that right? I can't understand how to find an expectation of $R_2$. 
