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.

Here is a sketch of the problem. I divided the $[0,1]\times[0,1]$ plane into 8 parts and shaded the areas where $\max(U,1/2)\leq X$. As you can see, it makes up $3/8$. (This is not a solution of course!)
Edit: corrected after feedback.

