Finding $P(X < Y)$ where $X$ and $Y$ are independent uniform random variables Suppose $X$ and $Y$ are two independent uniform variables in the intervals $(0,2)$ and $(1,3)$ respectively. I need to find $P(X < Y)$.
I've tried in this way:
$$
\begin{eqnarray}
P(X < Y) &=& \int_1^3 \left\{\int_0^y f_X(x) dx\right\}g_Y(y) dy\\
&=& \frac{1}{4} \int_1^3 \int_0^y dx dy\\
&=& \frac{1}{4} \int_1^3 y dy\\
&=& \frac{1}{8} [y^2]_1^3\\
&=& 1
\end{eqnarray}
$$
But I'm suspicious about this result. It implies that $X<Y$ is a sure event, which is not at all true. 
 A: $$\begin{align}
\mathsf P(X < Y) &=\int_1^3 \left\{\int_0^y f_X(x) \operatorname dx\right\}g_Y(y) \operatorname dy\\
&\neq \tfrac{1}{4} \int_1^3 \int_0^y \operatorname dx \operatorname dy
\end{align}$$
Here's the problem.   The inner integral's upper bound should be $\min(2, y)$ because the support for $X$ is $(0;2)$.   Watch out for this.
$$\begin{align}
\mathsf P(X < Y) &= \tfrac{1}{4} \int_1^3 \int_0^{\min(2,y)} \operatorname dx \operatorname dy
\\& = \tfrac 1 4 \left(\int_1^2\int_0^y\operatorname d x\operatorname d y
+ \int_2^3\int_0^2 \operatorname dx \operatorname d y\right)
\\& =\tfrac 1 4\left({\int_1^2 y\operatorname d y+\int_2^3 2 \operatorname d y }\right)
\\& =\tfrac 1 4\left(\tfrac 1 2(2^2-1^2)+ 2(3-2)\right)
\\& =\tfrac 7 8
\end{align}$$
That is all.
A: This sketch might help.  You want the red area as a proportion of the red and blue areas.

A: If you want to solve the problem using integrals then you should notice that you have wrong upper limit in the inner integral. It should be min(y,2).
A: We .can draw the rectangle and it's interior $ 0 \leq x \leq 2$ and $1 \leq y \leq 3$. Then we can draw the line $y=x$. Let's look at our event. So we should draw the line $y=x$. Now, the region delimited is given by the triangle whose vertices are $(1,1)$, $(2,2)$ and $(2,1)$.The probanility is $\int\limits_{1}^{2}\int\limits_{1}^{x} \frac{1}{4}dydx$.
A: Split the inteval and use Baye's Theorem to get
$$P(X<Y) = P(X<Y | X<1) P(X<1) + P(X<Y|X\geq1)P(X\geq 1))$$
sind $X$ is uniform on $(0,2)$, we know that $P(X<1)=P(X>1) = \frac12$. Furthermore, $P(X<Y|X<1) = 1$ since $Y$ is uniform on $(1,3)$. This yields
$$P(X<Y) = 1 \cdot \frac12  + P(X<Y|X\geq1)\frac12$$
Now $P(X<Y| X\geq 1) = P(X'<Y)$ where $X'$ is $\mathcal{U}(1,2)$. We can again split this and use Baye's to 
\begin{align}P(X'<Y) &= P(X'<Y| Y<2)P(Y<2) + P(X'<Y| Y\geq 2)P(Y\geq2)
\\
&= P(X'<Y|Y<2)\frac12 + 1\cdot \frac12
\end{align}
Now $P(X'<Y|Y<2) = P(X'<Y')$ with $Y'$ uniform in $(1,2)$. Since also $X$ is uniform in $(1,2)$, we have $P(X'<Y') = \frac12$.
Bringing everything together we obtain
\begin{align}
P(X<Y) = \frac12 + \left(\frac14+\frac12\right)\frac12 =\frac78
\end{align}
A: Answer:
Divide the regions of X with respect to Y for the condition $X<Y$.
For $0<X<1$, $P(X<Y) = \frac{1}{2}$
For $1<X<2$ and $1<Y<2$ $P(X<Y) = \int_{1}^{2}\int_{x}^{2} \frac{1}{2}\frac{1}{2}dydx = \frac{1}{8}$
For $1<X<2$, and $2<Y<3$ $P(X<Y) = \frac{1}{2}.\frac{1}{2}=\frac{1}{4}$
Thus $P(X<Y) = \frac{1}{2}+ \frac{1}{8}+\frac{1}{4} = \frac{7}{8}$
