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?
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 $Y=X-1$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
