# Probability of $p(x_1<x_2<\max(x_3,x_4))$

If $$x_1,x_2,x_3$$ 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 $$p(x_1 and every $$x$$ are iid so both have same probabiltiy so is $$2p(x_1 is right? But i get the $$p(x_1 and then my answer is $$1/3$$ but the answer is $$1/6.$$ Can we write $$P$$ as $$p(x_1?

• I did type-set as math, but what you write is not fully comprehensible. What you mean by iid? Should there be = before 1/6? Feb 17, 2020 at 15:39
• @emacsdrivesmenuts : "iid" is a standard and universally known abbreviation in probability theory. Feb 17, 2020 at 15:43

The two events $$x_1 < x_2 < x_3$$ and $$x_1 are not mutually exclusive, and the events $$x_1 and $$x_1 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 $$x_1 \begin{align} & x_1 < x_2 < x_3 < x_4 \\ & x_1 < x_2 < x_4 < x_3 \\ & x_1 < x_3 < x_2 < x_4 \\ & x_1 < x_4 < x_2 < x_3 \\ & x_3 < x_1 < x_2 < x_4 \\ & x_4 < x_1 < x_2 < x_3 \end{align} Since all $$24$$ orders are equally probable in this situation, the probability is therefore $$6/24 = 1/4 = 0.25.$$

• Can i answer this question by taking 1-p(max(x3,x4)<x2<x1) by breaking the event into two parts p(max(x3,x4)<x2) and p(x2<x1). Also, can anybody provide the answer by taking all mutually exclusive event of p(max(x3,x4)<x2<x1) May 6, 2020 at 19:50

$$p(x_1\lt x_2\lt\max(x_3,x_4))=p(x_1\lt x_2\lt x_3\text{ OR }x_1\lt x_2\lt x_4)$$ $$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\text{ AND }x_1\lt x_2\lt x_4)$$ $$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\lt x_4\text{ OR }x_1\lt x_2\lt x_4\lt x_3)$$ $$=\frac1{3!}+\frac1{3!}-\frac1{4!}-\frac1{4!}=\frac14.$$