# For continuous distribution $P(X_3<X_2<\max(X_1,X_4))$ equals? [duplicate]

Let $$X_1,X_2,X_3,X_4$$ be i.i.d random variables have a continuous distribution. Then $$P(X_3 equals?

$$A=\dfrac{1}{2}$$

$$B=\dfrac{1}{3}$$

$$C=\dfrac{1}{4}$$

$$D=\dfrac{1}{6}$$

My attempt:

$$P(X_3

I am blank now. No distribution is provided I don't know what I am missing here.

I know the fact that $$Z=F_{X}(x) \sim U(0,1)$$ but how do I use that here?

Ignoring equalities (which happen with probability $$0$$ since this is a continuous distribution) there are $$4!=24$$ equally likely orders

Of these, half $$(12)$$ have $$X_3>X_2$$ and can be excluded, while a quarter $$(6)$$ have $$X_2$$ as the largest value and can also be excluded.

Every other case has $$X_3 and so $$\mathbb P(X_3

The six orders which meet the condition are

• $$X_4 < X_3 < X_2 < X_1$$
• $$X_3 < X_4 < X_2 < X_1$$
• $$X_3 < X_2 < X_4 < X_1$$
• $$X_3 < X_2 < X_1 < X_4$$
• $$X_3 < X_1 < X_2 < X_4$$
• $$X_1 < X_3 < X_2 < X_4$$

so we can also say $$\mathbb P(X_3

• One thing how do I calculate the number of times my case matches? How did you get 6 ? I mean its count but how do I get it mathematically Jan 8, 2021 at 11:32
• @Daman $24-12-6=6.$ Jan 8, 2021 at 11:34