May 06'23
Exercise
A company offers a basic life insurance policy to its employees, as well as a supplemental life insurance policy. To purchase the supplemental policy, an employee must first purchase the basic policy. Let [math]X[/math] denote the proportion of employees who purchase the basic policy, and [math]Y[/math] the proportion of employees who purchase the supplemental policy. Let [math]X[/math] and [math]Y[/math] have the joint density function [math]f(x,y) = 2(x+y)[/math] on the region where the density is positive.
Given that 10% of the employees buy the basic policy, calculate the probability that fewer than 5% buy the supplemental policy.
- 0.010
- 0.013
- 0.108
- 0.417
- 0.500
May 06'23
Solution: D
We are given that the joint pdf of X and Y is f(x,y) = 2(x+y), 0 < y < x < 1 . Now
[[math]]
f_x(x) = \int_0^x (2x + 2y) dy = [ 2xy + y^2]_0^x = 2x^2 + x^2 = 3x^2, 0 \lt x \lt 1
[[/math]]
so
[[math]]
f(y|x) = \frac{f(x,y)}{f_x(x)} = \frac{2(x+y)}{3x^2} = \frac{2}{3}(\frac{1}{x} + \frac{y}{x^2}), 0 \lt y \lt x
[[/math]]
[[math]]
f(y|x = 0.10) = \frac{2}{3} [\frac{1}{0.1} + \frac{y}{0.1}] = \frac{2}{3}[10 + 100y], 0 \lt y \lt 0.10
[[/math]]
[[math]]
P[ Y \lt 0.05 | X = 0.10] = \int_0^{0.05} \frac{2}{3}[10 + 100y] dy = [ \frac{20}{3}y + \frac{100}{3}y^2]_0^{0.05} = \frac{1}{3} + \frac{1}{12} = \frac{5}{12} = 0.4167.
[[/math]]