Calculate the probability that a randomly chosen claim on this policy is processed in four hours or more I'm not sure how to approach this question, can someone please help me?
An insurance policy is written to cover a loss X where X has density function
f(x) = (2/9)x for 0 ≤ x ≤ 3
       and 0 otherwise
The time (in hours) to process a claim of size x, where 0 ≤ x ≤ 3, is uniformly distributed
on the interval x to 4x. Calculate the probability that a randomly chosen claim on this policy is processed in four hours or more.
A. 0.425
B. 0.593
C. 0.666
D. 0.721
E. None of the above
Ive attempted the double integral:
1 to 3 and 4 to 4x (2/27) dy dx
since f(x,y) is 2/27 (1/3x times (2/9)x)

Use the law of total probability. Let $T$ be the time it takes to process a claim. Then $T\mid X=x\sim \text{Unif}(x, 4x)$. Note that $P(T\geq 4\mid X=x)=\frac{4x-4}{3x}=\frac{4(x-1)}{3x}$ for $1\leq x\leq 3$ and equal to zero otherwise.Hence
$$
P(T\geq 4)=\int P(T\geq 4\mid X=x)f_{X}(x)\, dx=\int_{1}^3\frac{4(x-1)}{3x}\frac{2}{9}x\, dx
$$
which you can compute.
