I have solved the problem using definitions but I am not able to convince myself on one thing. Kindly if someone can throw some light to simplify the matter. Explanation required for the step between.
Problem
Given that $Y \sim \text{Uni}[0,4]$, $W \sim \text{Uni}[0,8]$ and $X$ is exponentially distributed with parameter $\lambda(y)=\frac{1}{5-y}$.
Find
i) $\mathbb{P}(W<Y)$, and
ii) $\mathbb{P}(W>X+Y)$.
Solution
i)
$$\mathbb{P}(W<Y)= \int\limits_0^4 \frac{1}{4} \int\limits_0^y \frac{1}{8} \,dw \,dx=\frac{1}{4}$$
I have added the picture of the region.
ii)
$$\mathbb{P}(W>X+Y)= \int\limits_0^4 \mathbb{P}(W>X+Y\mid Y=y)f_Y(y)\,dy $$ using total probability theorem
$f_{X\mid Y}(x\mid Y=y)=\frac{1}{5-y}e^{-\frac{x}{5-y}},\quad x\geq0$
$F_{X\mid Y}(x\mid Y=y)=1-e^{-\frac{x}{5-y}},\quad x\geq0$ and $0$ otherwise
$$\mathbb{P}(W>X+Y)= \int\limits_0^4 \mathbb{P}(X<W-Y\mid Y=y)f_Y(y)\,dy $$
$$\mathbb{P}(W>X+Y)=\int\limits_0^4 \int\limits_y^8 F_{X\mid Y}(w-y)f_W(w)f_Y(y)\,dw\,dy$$
I have used the fact
$$f_{S-U}(z)=\int\limits_{-\infty}^{\infty} f_S(s)f_U(s-z)\,ds$$
Need explanation for the above step, If possible some easy explanation with diagram. I assume the fact of fixing the value of $Y=y$ and then solving convolution of $W-X$, subtraction of the random variables.
$$\int\limits_0^4 \frac{1}{4} \int\limits_0^y \frac{1}{8} \int\limits_0^{w-y} \underbrace{\frac{1}{5-y}e^{-\frac{x}{5-y}}}_{f_{X\mid Y}(x\mid Y=y)} \,dx \,dw \,dy$$
$$\frac{12}{32} + \frac{1}{32} \int\limits_0^4 (5-y)e^{-\frac{8-y}{5-y}}\,dy $$
$$\frac{12}{32} + \frac{1.7584}{32}$$