<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] (See note below for the correct answer.)
Let $X_k = 1$ if the $k$th calculator is broken, and $0$ otherwise.
Assuming independence of 'brokenness', we have the probability of the $k$th calculator being broken is $P(X_k = 1) = \frac{20}{200} = \frac{1}{10}$.
Then the expected number of broken calculators is
$$E (\sum_{k=1}^{10} X_k) = \sum_{k=1}^{10}1 P(X_k) = 10 \frac{1}{10} = 1$$
NOTE: The above relies on the randomness of selection, not of any characteristic of 'brokenness'. Hence the reasoning is incorrect, even if the answer is correct. Here is a correct answer with correct reasoning.
Let $J = \{1,...,200\}$ represent the calculators. Let $B \subset J$, with $|B| = 20$, represent the (indices of) the broken calculators. Let $n = 10$ represent the number of calculators selected.
Our sample space is $\Omega = \{I \subset J | |I| = n \}$. Assuming the sample is selected uniformly, we have the probability $P(I) = \frac{1}{|\Omega|}$ for all $I \in \Omega$.
As above, let $X_k = 1$ if the $k$th calculator is broken, and $0$ otherwise, that is $X_k = 1_{B}(k)$. Obviously, we have $\sum_{k \in J} X_k = |B|$. Let $I \in \Omega$, then $\sum_{k \in I} X_k$ represents the number of broken calculators in the sample $I$.
We wish to compute $E (\sum_{k \in I} X_k)$, and since the distribution is uniform, we have $E (\sum_{k \in I} X_k) =  \sum_{I \in \Omega} \sum_{k \in I} X_k \frac{1}{|\Omega|}$.
To simplify this expression, count the number of times a given $X_k$ occurs in the above summation. There are a total of $n|\Omega|$ '$X_*$s, so any particular $X_k$ occurs $\frac{n|\Omega|}{|J|}$ times. Consequently we have 
$$E (\sum_{k \in I} X_k) =  \sum_{I \in \Omega} \sum_{k \in I} X_k \frac{1}{|\Omega|} = \frac{n|\Omega|}{|J|} \frac{1}{|\Omega|} \sum_{k \in J} X_k = \frac{n|B|}{|J|} = \frac{10\cdot 20}{200} = 1$$
</s>