$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.
Solution:
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $10$ calculators being broken, multiply the probabilities with the respective $0$ - $10$ and add those together?
For example:
\begin{align} {10 \choose 0} \cdot \left(\frac{1}{10}\right)^0 \cdot \left(\frac{9}{10}\right)^{10} &= 0.3487 \tag{0}, \\ {10 \choose 1} \cdot \left(\frac{1}{10}\right)^1 \cdot \left(\frac{9}{10}\right)^9 &= 0.3874 \tag{1}, \\ \vdots & \\ {10 \choose 10} \cdot \left(\frac{1}{10}\right)^{10} \cdot \left(\frac{9}{10}\right)^0 &= 1 \cdot 10^{-10}. \tag{10} \end{align}
Adding them together:
\begin{align} E[X] = (0) \cdot 0 + (1) \cdot 1 + \ldots + (10) \cdot 10. \end{align}
Or is it enough to calculate
\begin{align} E[X] &= (1+2+3+4+5+6+7+8+9+10) \cdot \left(\frac{1}{10}\right). \end{align}
Thanks.