Find $\Pr(Y\le 2X)$ for two iid random variables of a given pdf Consider two independent random variables $X$ and $Y$.
Let both PDFs are defined by $$f_X(x)=\begin{cases}\displaystyle\frac{4-2x}{3}&\text{if }x\in(0,1),\\0&\text{otherwise.}\end{cases}$$
Find $\Pr(Y \le 2X)$.

I solved this the following process.


*

*I've found
$$
\Pr(Y \le 2X | X=c) = \begin{cases}
0 & \text{if }c\le0\\
-\frac{4}{3}(c^2-2c) & \text{if } 0\lt c \lt \frac12\\
1 & \text{if }c\ge\frac12
\end{cases}
$$

*I've solved
\begin{align}
\Pr(Y \le 2X) &= \int_{-\infty}^{\infty} \Pr(Y\le 2X|X=x) f_X(x)~dx\\
&= \int_{0}^{1} \Pr(Y\le 2X|X=x) \cdot \frac{4-2x}{3} ~dx\\
&= \int_{0}^{\frac12} -\frac{4}{3}(x^2-2x) \cdot \frac{4-2x}{3} ~dx + \int_{\frac12}^1 1 \cdot \frac{4-2x}{3} ~dx\\
&= \frac{121}{216}
\end{align}



Is there anything wrong? Also, is there the better solution? Thank you for reading my question.

Yes, indeed, that's the right integral.  But not the right value.
$$\begin{align}\mathsf P(Y\leq 2X) ~=~& \int_0^{1/2}\frac{4-2x}3\int_{0}^{2x}\frac{4-2y}3\,\mathsf d y\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\int_0^1 \frac{4-2y}3\,\mathsf d y\,\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{4-2x}3\frac{4(2x-x^2)}3\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{32x-32x^2+8x^3}9\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\mathsf d x \\[1ex]=~& \frac 89\left[\frac{x^4}4-\frac{4x^3}3+2x^2\right]_{x=0}^{x=1/2}+\frac 13\left[4x-x^2\right]_{x=1/2}^{x=1} \\[2ex]=~& \dfrac{157}{261} \end{align}$$
