Find the probability of $X_2$ lying between $X_1$ and $X_3$ 
All $X_1,X_2,X_3$ are independent and uniformly distributed on
  $[0,1]$. Find the probability of $X_2$ lying between $X_1$ and $X_3$

Is the following method correct?
Find the $P(X_1<X_2<X_3) + P(X_3<X_2<X_1)$
Find $P(X_1<X_2<X_3)$
Fix $X_2$. Denote the point on $[0,1]$ as $x_2$. The probability of $P(X_1<X_2)$ is $x_2-0$ and the $P(X_2<X_3)=1-x_2$
$P(X_1<X_2<X_3)=(1-x_2)(x_2)$
 A: You can solve this problem by symmetry. The following events are distinct
$i$) $X_1$ lies between $X_2$ and $X_3$
$ii$) $X_2$ lies between $X_1$ and $X_3$
$iii$) $X_3$ lies between $X_1$ and $X_2$
and they are complete in the sense that one of them is always true, $P(i) + P(ii) + P(iii) =1$.
Because of the symmetry they are equivalent and we have $P(i) =P (ii)= P(iii)$  and thus
$$P(ii) = \frac13.$$
A: Each of the three values $X_1, X_2, X_3$ has  an equal chance of being the middle value.
A: You were close, but you just needed to integrate the variable over its support interval.
$$\begin{align}\mathsf P(X_1<X_2<X_3) & = \int_0^1 f_{X_2}(x_2)\; F_{X_1}(x_2)\; (1-F_{X_3}(x_2))\operatorname d x_2 \\[1ex] & = \int_0^1 1\cdot x_2\cdot(1-x_2)\operatorname d x_2
\\[1ex] & = {\left[\tfrac 1 2 {x_2}^2-\tfrac 1 3{x_2}^3 \right]}_{x_2=0}^{x_2=1}
\\[1ex] & = \tfrac 1 6
\end{align}$$
And likewise $\mathsf P(X_3<X_2<X_1) =\tfrac 1 6$  so the probability that $X_2$ lies between $X_1$ and $X_3$ is $\tfrac 1 3$; as we should anticipate by symmetry -- we have three independent, identically distributed, continuous random variables , so any one of them will be between the other two with equal probability.
