$X\sim U(0,1)$, and conditional distribution of $Y$ given $X$ is $U(0,x)$. Find $P(X+Y\geq1)$

Question

The question states:

Let $X$ have uniform distribution $U(0,1)$, and let the conditional distribution of $Y$, given $X=x$, be $U(0,x)$. Find $P(X+Y\geq1)$.

My attempt:

$f_X(x)=1, 0\leq x\leq 1$

$f_{Y|X}(y|x)=\frac{1}{x}, 0\leq y\leq x$

$f_{XY}(x,y)=\frac{1}{x}$

$P(X+Y)\geq 1) = 1 - P(X+Y<1)$

$$=1-\int_{0}^{1/2}\int_{y}^{1-y} \frac{1}{x} dx dy =1-\ln(2)$$

Is this right?

Answer 1

Seems correct. Integrating in the other order as a sanity check:

$$P(X+Y < 1) = \int_0^{1/2} \int_0^x \frac{1}{x} \, dy \, dx + \int_{1/2}^1 \int_0^{1-x} \frac{1}{x} \, dy \,dx = \frac{1}{2} + \int_{1/2}^1 \left(\frac{1}{x} - 1\right) \, dx = \ln 2.$$