Expectation of joint life span The life span of a particular mechanical part is a random variable described by the following PDF:

If three such parts are put into service independently at t=0, determine a simle expression for the expected value of the time until the majority of the parts will have failed.
I can get the PDF:
$$
f_L(l) = 0.4 (0 \leq l \leq 2) \\
f_L(l) = -0.4l + 1.2 (2 < l \leq 3)
$$
and the expectation:
$$
E(l) = \int_0^3 l f_L(l) dl \approx 1.27
$$
I think 'majority' means 2 or more, so we can focus on two parts of the three, and pay no attention to the third. The translation is $E(max(l1, l2))$, how will this be derived I currently have no idea.

Sorry about the misleading remark "$E(max(l_1, l_2))$", it's wrong to neglect the third part, because if that one fails early, then we only need one of the rest to fail.
 A: Let $X$ denote the life span of any given component and $T$ the first time when at least 2 out of 3 components fail. The event $[T\gt t]$ means either that none of the 3 components fails before time $t$ or that exactly 1 component out of 3 fails before that time, hence, for every $t\gt0$,
$$
\mathbb P(T\gt t)=\mathbb P(X\gt t)^3+3\mathbb P(X\gt t)^2\mathbb P(X\lt t),
$$
that is,
$$
\mathbb P(T\gt t)=1-3u(t)^2+2u(t)^3=v(t)^2(3-2v(t)),
$$
with
$$
u(t)=\mathbb P(X\lt t),\qquad v(t)=1-u(t)=\mathbb P(X\gt t).
$$
Furthermore,
$$
\mathbb E(T)=\int_0^{+\infty}\mathbb P(T\gt t)\mathrm dt.
$$
In the present case, the density of $X$ is $f_X(t)=\frac25$ if $0\lt t\lt 2$ and $f_X(3-t)=\frac25t$ if $0\lt t\lt 1$. Hence
$u(t)=\frac25t$ if $0\lt t\lt 2$ and $v(3-t)=\frac15t^2$ if $0\lt t\lt 1$. This yields
$$
\mathbb E(T)=\int_0^2(1-3u(t)^2+2u(t)^3)\mathrm dt+\int_0^1v(3-t)^2(3-2v(3-t))\mathrm dt,
$$
that is,
$$
\mathbb E(T)=\int_0^2(1-\tfrac{12}{25}t^2+\tfrac{16}{125}t^3)\mathrm dt+\int_0^1\tfrac1{25}t^4(3-\tfrac25t^2)\mathrm dt,
$$
or,
$$
\mathbb E(T)=\left[t-\tfrac{4}{25}t^3+\tfrac{4}{125}t^4\right]_{t=0}^{t=2}+\left[\tfrac3{125}t^5-\tfrac2{125}\tfrac17t^7\right]_{t=0}^{t=1},
$$
that is,
$$
\mathbb E(T)=2-\tfrac{32}{25}+\tfrac{64}{125}+\tfrac3{125}-\tfrac2{125\cdot7}=\tfrac{1097}{875}=1.253\overline{714285}.
$$
A: The value for $E \left[ \max \left( L_1, L_2 \right) \right]$ is computed in
the following way. First, the distribution of the maximum of two identically
independently distribued random variable $L_1$ and $L_2$ is given by $2 f
\left( \ell \right) F \left( \ell \right)$ where $f \left( \ell \right)$ is
the density and $F \left( \ell \right)$ is the cumulative distribution
function. This is well known, you could find the formula here. It is not
difficult to derive:
\begin{eqnarray*} 
  \Pr \left[ \max \left( L_1, L_2 \right) \leqslant \ell \right] & = & \Pr
  \left[ \left\{ L_1 \leqslant \ell \right\} \cap \left\{ L_2 \leqslant \ell
  \right\} \right]\\
  & = & \Pr \left[ L_1 \leqslant \ell \right] \Pr \left[ L_2 \leqslant \ell
  \right]\\
  & = & F \left( \ell \right)^2
\end{eqnarray*}
Taking derivative gives the density $2 f \left( \ell \right) F \left( \ell
\right)$.
The probability density function $f(\ell)$ is given by (as you indicated)
$$ f \left( \ell \right) = \frac{2}{5} 1_{\ell} \left[ 0, 2 \right) + \left(
   - \frac{2}{5} \ell + \frac{6}{5} \right) 1_{\ell} \left[ 2, 3 \right) $$
where the notation $1_{\ell}A$ with interval $A$ is that of an indicator variable. This means
$$ 1_{\ell} \left( A \right) = \left\{ \begin{array}{lll}
     1 &  & \text{if } \ell \in A\\
     0 &  & \text{otherwise}
   \end{array} \right. $$
Therefore the cumlative distribution function is given by
$$ F( \ell)= \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2
   \right) + \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[
   2, 3 \right) + 1_{\ell} \left[ 3, \infty \right) $$
Multiplying both we get the density
\begin{eqnarray*}
  2 f \left( \ell \right) F \left( \ell \right) & = & 2 \left\{ \frac{2}{5}
  1_{\ell} \left[ 0, 2 \right) + \left( - \frac{2}{5} \ell + \frac{6}{5}
  \right) 1_{\ell} \left[ 2, 3 \right) \right\}\\
  & \times & \left\{ \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2 \right) +
  \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3
  \right) + 1_{\ell} \left[ 3, \infty \right) \right\}\\
  & = & \frac{8 \ell}{25} 1_{\ell} \left[ 0, 2 \right) + \frac{4}{25} \left(
  - \ell + 3 \right) \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3
  \right)
\end{eqnarray*}
and therefore
\begin{eqnarray*}
  E \left[ \max \left( L_1, L_2 \right) \right] & = & \frac{8}{25} \int_0^2
  \ell^2 \mathrm{d} \ell + \frac{4}{25} \int_2^3 \ell \left( - \ell + 3
  \right) \left( - \ell^2 + 6 \ell - 4 \right) \mathrm{d} \ell\\
  & = & \frac{637}{375}\\ 
 & \approx & 1.69867
\end{eqnarray*}
