Expected value binomial proof A box contains 10 pens, and the probability that any given pen is defective is $p$.
The expected value is $\sum_{X = 0}^{10} X \binom{10}{X}p^X(1-p)^{10-X}$.
Using the binomial theorem, I know that $\sum_{X = 0}^{10} \binom{10}{X}p^X(1-p)^{10-X} = 1$.
By testing different values of $p$, I found that the expected value is $10p$.
Is there a way of proving this?
 A: The expected value is $\sum_{X=0}^{10}\binom{10}{X}Xp^{X}(1-p)^{10-X}=10p\sum_{Y=0}^9\binom{9}{Y}p^Y(1-p)^{9-Y}=10p$. by letting $Y=X-1$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{X = 0}^{10}X{10 \choose X}p^{X}
\pars{1 - p}^{10 - X}} =
\left.\pars{1 - p}^{10}\sum_{X = 0}^{10}X
{10 \choose X}\alpha^{X}
\,\right\vert_{\ \alpha\ =\ p/\pars{1 - p}}
\\[5mm] = &\
\left.\pars{1 - p}^{10}\,\alpha\,\partiald{}{\alpha}\sum_{X = 0}^{10}{10 \choose X}\alpha^{X}
\,\right\vert_{\ \alpha\ =\ p/\pars{1 - p}}
\\[5mm] = &\
\left.\pars{1 - p}^{10}\,{p \over 1 - p}\,\partiald{\pars{1 + \alpha}^{10}}{\alpha}
\,\right\vert_{\ \alpha\ =\ p/\pars{1 - p}}
\\[5mm] = &\
\pars{1 - p}^{9}\, p\bracks{10\pars{1 + {p \over 1 -p}}^{9}} =
\bbx{10p}
\end{align}
