I'm confused about how to interpret the given conditions of this problem.
"An insurance policy pays for a random loss $X$ subject to a deductible of $C$ where $0<C<1$. The loss amount is modeled as a continuous random variable with density function $$f(x) = \begin{cases}2x~~ 0<x<1 \\ 0~~\text{otherwise}\end{cases}$$ $\textbf{Given a random loss $X$}$ the probability that the insurance payment is less than .5 is .64. Calculate $C$."
The phrase "given a random loss $X$" suggested to me that .64 was equal to a conditional probability involving $X$. But according to the posted solution, $C=.3$ is obtained by setting the unconditioned probability $P(Y\leq .5)$ equal to .64 (where $Y = $ insurance payment = $X-C$) and then solving for $C$ using $$\int_{0}^{.5+C} 2x~dx = .64$$
My question is, what happened to the condition? Is "given a loss $X$" already baked into the expression $P(X-C\leq .5)$, and therefore just a bit of misdirection? My own attempt involved defining $$Y= \begin{cases} 0~~\text{if } X\leq C\\\ X-C~~ \text{if }X>C\end{cases}$$
and then solving for $C$ using$$P(Y\leq .5\vert X>C) = \frac{P(X-C \leq .5 \cap X>C)}{P(X>C)} = \frac{P(C \leq X \leq C+.5)}{1-P(X\leq C)} = \frac{\int_{C}^{C+.5} 2x~dx}{1 - \int_{0}^{C} 2x~dx} = .64$$ That gives $C \approx .32316$
But that didn't seem right because it skips over the losses that are less than $C$. Any help with answering these questions would be greatly appreciated!
