Finding the expected value in a random drawing 
Suppose the life span of a battery follows exponential distribution with the expected value of 1 year. If three battery are drawn at random and tested until they all fail, what is the expected life span of the battery that failed second?

(There is a duplicate of this question but it's on hold so I'm posting another one)
I'm thinking about setting the expected value of the three batteries as a random vector, say ($v_1, v_2, v_3$). So the one that failed the last would be max$(E(v_1),E(v_2),E(v_3))$. But I'm not sure that if this approach is correct,  and if it is, how should I proceed to find the middle value.
 A: If we let $X_j$ be the lifetime of battery $j$, then $X_j \sim Exp(\lambda)$ i.i.d., where $\frac{1}{\lambda} = 1 \text{ year}$.
One could then consider the order statistics $X_{(1)}$, $X_{(2)}$ and $X_{(3)}$, which are the ordered lifetimes of the batteries, so $X_{(1)} \le X_{(2)} \le X_{(3)}$. We thus need to compute $E[ X_{(2)}]$.
Using https://en.wikipedia.org/wiki/Order_statistic#Order_statistics_sampled_from_an_exponential_distribution, the random variable $X_{(2)}$ can be written as
\begin{equation}
X_{(2)} = \frac{1}{\lambda}\left( \frac{Z_1}{3} + \frac{Z_2}{2} \right),
\end{equation}
where $Z_1$ and $Z_2$ are exponential random variables with parameter $1$.
The expected value is then
\begin{align}
E \left[ X_{(2)} \right] &= E \left[\frac{1}{\lambda}\left( \frac{Z_1}{3} + \frac{Z_2}{2} \right)\right] \\
&= \frac{1}{\lambda}\left( E \left[ \frac{Z_1}{3} \right] + E \left[ \frac{Z_2}{2} \right] \right) \\
&= \frac{1}{\lambda}\left( \frac{1}{3} + \frac{1}{2} \right) \\
&= \frac{5}{6}\cdot \frac{1}{\lambda}.
\end{align}
Therefore, the expected time of failure of the battery failing second is $\frac{5}{6}$ years.
