# Exam P Practice Problem 104 – two random insurance losses

Problem 104-A

Two random losses $X$ and $Y$ are jointly modeled by the following density function: $\displaystyle f(x,y)=\frac{1}{32} \ (4-x) \ (4-y) \ \ \ \ \ \ 0

Suppose that both of these losses had occurred. Given that $X$ is exactly 2, what is the probability that $Y$ is less than 1? $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{7}{24}$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{11}{24}$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{12}{24}$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{24}$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{24}$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$

Problem 104-B

Two random losses $X$ and $Y$ are jointly modeled by the following density function: $\displaystyle f(x,y)=\frac{1}{96} \ (x+2y) \ \ \ \ \ \ 0

Suppose that both of these losses had occurred. Determine the probability that $Y$ exceeds 2 given that the loss $X$ is known to be 2. $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{36}$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{24}{36}$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{26}{36}$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{28}{36}$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{29}{36}$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$

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