$\operatorname E\left( \min \left( X_1,X_2\right) \right) $ where $X_1\sim U( 0,5) $ and $X_2\sim U(0,3)$ X1, and X2 are independent
$$\operatorname E\left( \min \left( X_1,X_2\right) \right) \text{ where } X_1\sim U\left( 0,5\right) \text{ and } X_2\sim U\left( 0,3\right)$$
let me label $\min(X_1,X_2) = X$
Or to make the problem more fun:  You enter a metro station in a big hurry and decide to take the first train that arrives. There are two lines running through this station: one runs every five minutes (line A), the other every three (line B). They are uniformly distributed, one over $[0,3]$ and the other over $[0,5]$
How many minutes will you wait on average until you get on a train?
Now my attempt, I feel like I have fortuitously stumbled on the solution with the wrong method.
\begin{align}
& \operatorname E(\min(X_1,X_2)) = \operatorname E( X\mid X_1>3 ) P(X_1 > 3) +  E(X\mid X_1<3 ) P(X_1 < 3) \\[10pt]
= {} & \frac 3 2 \cdot \frac 2 5 + \frac 3 5\operatorname E(X\mid X_1<3
\end{align}
I think this is correct it is just calculating $\operatorname E(X\mid X_1<3) $, which I am not sure how to go about.
I did $\operatorname E(X\mid X_1<3) = \int^3_0 \left( 1-\dfrac {x}{3}\right)^2 \, dx=1,$ which gives the right answer,, but  I am not sure if it is right, as this is just $\int^3_0 p(x) \, dx$  
Is it just lucky it gave the answer or does $E(X\mid X_1<3) = \int ^{3}_{0}P\left( x\geq x | X_1<3\right) dx$= $\int^3_0 \left( 1-\dfrac {x}{3}\right)^2 \, dx=1$

Even though you have not stated anything about the independence of $X_1$ and $X_2$, I am assuming they are independent. It is also quite trivial that the "lines" should be independent of each other.  
Before proceeding to expectation, we require p.d.f of $Y=min(X_1,X_2)$
Suppose $F_{Y}(y)$ be the c.d.f of $Y$.
$F_{Y}(y)=P(Y\le y)=1-P(Y > y)=1-P(X_1>y,X_2>y)$
$=1-[P(X_1>y)\times P(X_2>y)]$ ($X_1,X_2$ being independent)
$=1-[\int_y^5 \frac{1}{5} dx_1 \times \int_y^3 \frac{1}{3} dx_2]$ (The integrals are valid only when $y \le 3 $)
$=\frac{8y-y^2}{15}$   
Differentiating the relation, we obtain the p.d.f of $Y$ as  $f_Y(y)=\frac{8-2y}{15},0 <y<3 $
($y$ can't exceed $3$)  
Now, $E(min(X_1,X_2))=E(Y)=\int_0^3 y f_Y(y) dy=\int_0^3 y (\frac{8-2y}{15} )dy=\frac{18}{15}=\frac{6}{5}$  
The expected value of $Y$ is greater than $1$
