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Let $X_1, X_2$, and $Y$ be independent uniform random variables on (0,1).

For some reason I'm getting stuck on this, and I can't figure out why...

$\mathbb{P}[(Y<X_2)\cap (Y>X_1)] = $?

So to me, this looks like it can be written as $\mathbb{P}[X_1<Y<X_2]$, which yields

\begin{align*} &\int_0^1\int_0^1\int_0^1 \mathbb{1}_{(x_1<Y<x_2)} dy \,dx_1 \,dx_2\\ &\int_0^1\int_0^1\int_{x_1}^{x_2} dy \,dx_1 \,dx_2 = \int_0^1\int_0^1(x_1 - x_2) \,dx_1 \,dx_2 = \int_0^1 \tfrac{1}{2} - x_2 \,dx_2 = 0? \end{align*}

That doesn't seem right, but I can't see it.

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\begin{align} & P[Y < X_2, Y > X_1] \\ = & \int_{(0, 1)}P[X_2 > y, X_1 < y]f_Y(y)dy \\ = & \int_0^1 P[X_2 > y]P[X_1 < y]dy \\ = & \int_0^1 (1 - y)ydy \\ = &\frac{1}{2} - \frac{1}{3} \\ = & \frac{1}{6}. \end{align}

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This is equal to the probability that we have $X_1<Y<X_2$. Note that if if you picked three numbers at random and then assigned these numbers randomly to $X_1,X_2$ and $Y$ that the probability the numbers were assigned in increasing order is $1/6$. There are $6$ ways of assigning the numbers and only one way is strictly increasing.

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