Probability of Uniform distribution of a random variable $X$ when minimum value of X is involved

Question

If $X$ is a $U(0,1)$ random variable, then $P(min(X,1-X)\leq \frac{1}{4})$ =

I know the value of PDF will be $1$ as b and a is $0$ and $1$.

I dont know how to do when $min$ is involved.

How do we approach this question?

Answer 1

$P(\min (X,1-X) \leq \frac 14)=P(X \leq \frac 1 4 \vee X\geq \frac 34)=\frac 14+(1-\frac 3 4)=\frac 1 2$.