The expected number of heads Given $10$ fair coins:

*

*In the first round, we toss each coin once which gives us a combination of heads and tails.

*In the second round, we only toss those coins that landed on the tail in the first round.

What is the expected number of heads after this experiment $?$
Intuition tells me is $5 + 2.5 = 7.5$.
 A: An alternate solution is as follows: imagine you flipped all the coins twice.  Then any coin that gave you heads on the first flip or the second flip would be one of the ones you want to count.  The probability of getting at least one head in two flips is $3/4$, so the expected number of coins that get at least one head is $10 \times 3/4 = 7.5$.
A: Intuition is not enough to solve an exercise...
The expected #H in the first round is evidently $10\times\frac{1}{2}=5$
In the second round,
$Y|X\sim Bin\Big(10-x;\frac{1}{2}\Big)$
thus
$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}\Bigg[\frac{10-x}{2}\Bigg]=5-\frac{1}{2}\mathbb{E}[X]=\frac{5}{2}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\overbrace{\color{#f44}{\sum_{k = 0}^{n}{n \choose k}\pars{1 \over 2}^{k}
\pars{1 \over 2}^{n - k}}}
^{\substack{\ds{First\ Round:}\\[1mm] \ds{\color{#f44}{k}\ \mbox{heads}}}}\,\,\,
\overbrace{\color{#44f}{\sum_{j = 0}^{n - k}{n - k \choose j}\pars{1 \over 2}^{j}
\pars{1 \over 2}^{n - k - j}}}
^{\substack{\ds{Second\ Round:} \\[1mm] \ds{\color{#44f}{j}\ \mbox{heads}}}}
\\[2mm] &\ \times\pars{k + j}
\\[5mm] = &\
{1 \over 2^{2n}}\sum_{k = 0}^{n}\sum_{j = 0}^{n - k}{n \choose k}
{n - k \choose j}2^{k}\pars{k + j} = \bbx{{3 \over 4}\,n}
\\[5mm] &\ \stackrel{\ds{n\ =\ 10}}{\ds{\implies}}\quad\bbx{7.5} \\ &
\end{align}
