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.
 A: Let's first do it geometrically and then see if that leads to an algebraic insight.
Let $X$ and $U$ be plotted on a coordinate plane, with the $X$-value horizontal and the $U$-value vertical.  Then $(X,U)$ is a point in the unit square.  If $U \le 1/2$, then $\max(U, 1/2) = 1/2$, and we compare $1/2$ against $X$, meaning that in the lower half of the unit square, the set of points $(X,U)$ satisfying the criterion $\max(U, 1/2) \le X$ comprises a square with vertices $(1/2, 0), (1, 0), (1, 1/2), (1/2, 1/2)$, since any $X \in [1/2, 1]$ will be at least as large as $\max(U, 1/2) = 1/2$.
What happens in the upper half of the square?  In this case, $\max(U, 1/2) = U$, and we require $X$ to be at least as large as $U$, so this includes all the points on or below the line $U = X$ in this upper half; i.e., it is the triangle with vertices $(1/2, 1/2), (1, 1/2), (1,1)$.
Combined together, these two regions become a trapezoid which is easily seen to have area $3/8$.
How do we use this to reason algebraically?  We simply write
$$\begin{align}
\Pr[\max(U, 1/2) \le X] 
&= \Pr[(1/2 \le X) \cap (U \le 1/2)] + \Pr[(U \le X) \cap (U > 1/2)] \\
&= \Pr[1/2 \le X]\Pr[U \le 1/2] + \Pr[1/2 < U \le X] \\
&= (1/2)(1/2) + (1/8) \\
&= 3/8.
\end{align}$$
If $\Pr[1/2 < U < X]$ is not obvious, we can always integrate:
$$\Pr[1/2 < U < X] = \int_{u=1/2}^1 \int_{x=u}^1 \, dx \, du = \int_{u=1/2}^1 1-u \, du = \left[u - \frac{u^2}{2}\right]_{u=1/2}^1 = \frac{1}{2} - \frac{3}{8} = \frac{1}{8}.$$
A: Your step at * is incorrect since the intersection is of two events that are not independent (both involve $X$).  Such issues can be dealt with by the law of total probability, but it can help to visualize it first.
It can be visualized by plotting the region where $\max\{U,1/2\}<X$ in the $(U,X)$ plane within the unit square (it will be a trapezoidal region) and the result should be apparent.
Mathematically, use the law of total probability:
$\mathbb{P}(\max\{U,1/2\}<X)=\mathbb{P}(\max\{U,1/2\}<X|U<1/2)\mathbb{P}(U<1/2)+\mathbb{P}(\max\{U,1/2\}<X|U \geq 1/2)\mathbb{P}(U \geq 1/2)$.
Note that $\mathbb{P}(X> U|U\geq 1/2)=1/4$ (visualize this or compute a double integral). You can finish the rest.
A: 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.

A: 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:
$$\begin{align*}
\int_{[0,1]}\int_{[0,1]}1_{u\le x}1_{\frac{1}{2}\le x}du\ dx &=\int_{[0,1]}\Big(\int_{[0,1]}1_{u\le x}du\Big)1_{\frac{1}{2}\le x}dx\\
&=\int_{[0,1]}\Big(\int_{0}^x du\Big)1_{\frac{1}{2}\le x}dx\\
&=\int_{[0,1]}x1_{\frac{1}{2}\le x}dx\\
&=\int_{\frac{1}{2}}^1xdx\\
&=\frac{1}{2}x^2\bigg|_{x=\frac{1}{2}}^{x=1}\\
&=\frac{1}{2}-\frac{1}{8}\\
&=\frac{3}{8}
\end{align*}$$
A: Breaking in (*)
$$\Pr[\max(U, 1/2) \le X]=\Pr\left( U  \leq X \right) \cdot \Pr\left( \frac{1}{2} \leq X\right)$$
is not correct because both events depend on $X$. The closest proper reasoning is to condition on $X$:
$$\Pr[\max(U, 1/2) \le X]=\mathbb E^X[\Pr(\max(U, 1/2) \le X|X)]=\mathbb E^X[\Pr\left( U  \leq X | X\right) \mathbb{I}_{\frac{1}{2} \leq X}]=\mathbb E^X[X\mathbb{I}_{\frac{1}{2} \leq X}]$$
