Calculating probabilities in two overlapping continuous uniform random variables (with an added constraint)

Question

Given X~unif(a, b) and Y~unif(c, d) with a < c < b < d.

What's the probability that Y>X and Y being realized in the interval (c, b)?

Answer 1

Here we assume that $X$ and $Y$ are independent random variables with uniform distributions as specified by OP.

Write out the definition of the probability: $$ \mathsf{Pr}\left(Y>X, c<Y<b\right) = \int_{c}^{b} \left( \int_{a}^{y}\frac{\mathrm{d}x}{b-a} \right) \frac{\mathrm{d}y}{d-c} $$ Can you finish it off?

Answer 2

Here is one solution without explicit integration. Let $A$ be the event that $Y>X$ and $B$ the event that $c<Y<b$. Then we know that $\mathrm{Pr}(A \cap B)=\mathrm{Pr}(A|B)\mathrm{Pr}(B)$. The two probabilities on the right are simpler to calculate. First $\mathrm{Pr}(B)=\frac{b-c}{d-c}$.

Next we get a handle on $\mathrm{Pr}(A|B)$. Split the event $A$ into two disjoint events: $A_1$ and $A_2$ where $A_1$ is the event that $Y>X$ and $X<c$; and $A_2$ is the event that $Y>X$ and $X>c$. Then $\mathrm{Pr}(A_1)=\mathrm{Pr}(X<c)=\frac{c-a}{b-a}$ and $\mathrm{Pr}(A_2)=\frac{1}{2}\mathrm{Pr}(X>c)=\frac{1}{2}\left(\frac{b-c}{b-a}\right)$. Putting these two together, we have that $\mathrm{Pr}(A|B)=\frac{c-a}{b-a}+\frac{1}{2}\left(\frac{b-c}{b-a}\right)$.

Finally, using $\mathrm{Pr}(A \cap B)=\mathrm{Pr}(A|B)\mathrm{Pr}(B)$ and doing some algebra we find that $\mathrm{Pr}(A \cap B)=\frac{-2ab+b^2+2ac-c^2}{2(a-b)(c-d)}$.