Probability of $p(x_1If $x_1,x_2,x_3$ and $x_4$ are independent identically distributed  random variable with continuous distribution .what is the probability of
$$P=p(x_1 < x_2< \max(x_3,x_4))$$
I know the exhaustive number of cases are $p(x_1<x_2<x_3)+p(x_1<x_2<x_4)$ and  every $x$ are iid so both have same probabiltiy so is $2p(x_1<x_2<x_3)$ is right? But  i get the $p(x_1<x_2<x_3) = 1/6$ and then my answer is $1/3$ but the answer is $1/6.$ Can we write $P$ as $p(x_1<x_2<x_3<x_4)+p(x_1<x_2<x_4<x_3)$?
 A: The two events $x_1 < x_2 < x_3$ and $x_1<x_2<x_4$ are not mutually exclusive, and the events $x_1<x_2<x_3<x_4$ and $x_1<x_2<x_4<x_3$ are not exhaustive. That is where your attempts at solution go wrong.
One can look at it like this:
\begin{align}
& x_1 < x_2 < \max
\end{align}


*

*Either $x_3$ or $x_4$ could be $\max,$

*and then the one of those two that is not $\max$ could be greater than $x_2$ or between $x_2$ and $x_1$ or less than $x_1.$
I.e. choose one of two alternatives, then one of three. So there are six possibilities.
So six different orders favor the event $x_1<x_2 < \max\{x_3,x_4\}{:}$
\begin{align}
& x_1 < x_2 < x_3 < x_4 \\
& x_1 < x_2 < x_4 < x_3 \\
& x_1 < x_3 < x_2 < x_4 \\
& x_1 < x_4 < x_2 < x_3 \\
& x_3 < x_1 < x_2 < x_4 \\
& x_4 < x_1 < x_2 < x_3 
\end{align}
Since all $24$ orders are equally probable in this situation, the probability is therefore $6/24 = 1/4 = 0.25.$
A: $$p(x_1\lt x_2\lt\max(x_3,x_4))=p(x_1\lt x_2\lt x_3\text{ OR }x_1\lt x_2\lt x_4)$$
$$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\text{ AND }x_1\lt x_2\lt x_4)$$
$$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\lt x_4\text{ OR }x_1\lt x_2\lt x_4\lt x_3)$$
$$=\frac1{3!}+\frac1{3!}-\frac1{4!}-\frac1{4!}=\frac14.$$
