<s>[INST] Expected Number of Successes in a Sample 
$200$ 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.
 [/INST] The answer should be
$\left({20 \choose 1}{180 \choose 9} + 2{20 \choose 2}{180 \choose 8} + \cdots
+ 10{20 \choose 10}{180 \choose 0} \right) / {200 \choose 10}$, where ${n \choose k}$ denotes the binomial coefficient ${n \choose k} = \frac{n!}{k!(n-k)!}$.
This follows because the probability $p_k$ that you your sample has $k$ broken calculators is ${20 \choose k}{180 \choose 10-k} / {200 \choose 10}$. Why? Because the denominator is counting the total number of ways you could sample $10$ calculators from a set of $200$, and the numerator is counting the total number of such subsets which have $k$ broken calculators, because it's equivalent to sampling $k$ calculators from the set of broken ones and $10-k$ calculators from the set of non-broken ones.
The expected value is therefor $p_1 + 2p_2 + 3p_3 + \cdots + 10p_{10}$ which is equal to the number I have on top.
</s>