Probability of $P\left(X \in \left[\frac{a + 3b}{4}, b \right] \right)$ for uniform distribution

Question

For $X ∼ U(a,b)$, with $a,b > 0$, what is $P\left(X \in \left[\frac{a + 3b}{4}, b \right] \right)$?

I do not know how to solve this.

Does $$P(X \in [(a + 3b)/4, b]) = P((a + 3b)/4 < X < b)~?$$

Any help is appreciated. Thanks!

Answer 1

HINT: For $X \sim U([a,b])$, the probability that $\Pr(\alpha < X < \beta)$ is $$ \Pr(\alpha < X < \beta) = \int\limits_{\max(\alpha,a)}^{\min(\beta,b)} \frac{1}{b-a} \ dx$$ for $a<b$ and $\alpha<\beta$.