SOA Exam P Question 145 | Conditional Joint Variance

New dental and medical plan options will be offered to state employees next year. An
actuary uses the following density function to model the joint distribution of the
proportion X of state employees who will choose Dental Option 1 and the proportion Y
who will choose Medical Option 1 under the new plan options:
Calculate Var (Y | X = 0.75).
(A) 0.000
(B) 0.061
(C) 0.076
(D) 0.083
(E) 0.141

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[matthewgeary]

I don't understand how this question makes any sense. f(x, y) is a joint density function, as per the question, and thus must adhere to the Law of Total Probability stating that all valid probabilities must fall within the range [0, 1]. Thus, how can f(x, y) be valued at 1.25 for x ∈ [0, 0.5], y ∈ [0.5, 1] and 1.5 for x ∈ [0.5, 1], y ∈ [0, 0.5]? This is effectively saying that the probability that the proportion X choosing dental option 1 falling between 0% and 50% and the proportion Y choosing medical option 1 falling between 50% and 100% combined has a 125% chance of occurring, which not only is invalid on its own but would also invalidate the rest of the density function because that would imply that it is statistically impossible for X and Y to fall outside of those values.