Probability calculation for two independent uniform random variables I want to calculate the probability $\mathbb{P}\left( \max \{ U, \frac{1}{2} \} \leq X\right) $ with $U, X \sim $Unif$[0,1]$ and independent. I know that the result is $\frac{3}{8}$, but do not really know how to get there. I tried
\begin{align*}
\mathbb{P}\left( \max \{ U, \frac{1}{2} \} \leq X\right) = \mathbb{P}\left( U  \leq X \text{ and } \frac{1}{2} \leq X\right) \overset{(*)}{=} \underbrace{\mathbb{P}\left( U  \leq X \right)}_{= \frac{1}{2}} \cdot \underbrace{\mathbb{P}\left( \frac{1}{2} \leq X\right)}_{= \frac{1}{2}} = \frac{1}{4}.
\end{align*}
At (*) I used the Independence of $X$ and $U$. Obviously there must be a mistake at some point. Can anybody tell me how to get to $\frac{3}{8}$? It can't be that hard, but right now I do not know how to do it properly.

Use the joint density function:
$$P(max(U,\frac{1}{2})\le X)=\int_{[0,1]\times[0,1]} 1_{max(u,\frac{1}{2})\le x}d(u,x)=\int_{[0,1]\times[0,1]} 1_{u\le x}1_{\frac{1}{2}\le x}d(u,x)$$
Now, since $U$ and $X$ are independent, it holds
$$\int_{[0,1]\times[0,1]} 1_{u\le x}1_{\frac{1}{2}\le x}d(u,x)=\int_{[0,1]}\int_{[0,1]}1_{u\le x}1_{\frac{1}{2}\le x}du\ dx$$
This is calculated as follows:
$$\begin{align*}
\int_{[0,1]}\int_{[0,1]}1_{u\le x}1_{\frac{1}{2}\le x}du\ dx &=\int_{[0,1]}\Big(\int_{[0,1]}1_{u\le x}du\Big)1_{\frac{1}{2}\le x}dx\\
&=\int_{[0,1]}\Big(\int_{0}^x du\Big)1_{\frac{1}{2}\le x}dx\\
&=\int_{[0,1]}x1_{\frac{1}{2}\le x}dx\\
&=\int_{\frac{1}{2}}^1xdx\\
&=\frac{1}{2}x^2\bigg|_{x=\frac{1}{2}}^{x=1}\\
&=\frac{1}{2}-\frac{1}{8}\\
&=\frac{3}{8}
\end{align*}$$
