Question

The loss due to an earthquake in a commercial building is modelled by a random
variable X with density function
f(x) =  0:005(20 0; − x); for 0 elsewhere < x < 20
Given that the fire loss exceeds 10, what is the probability that it exceeds 18

Answer 1

$P(X>10)=\int_{10}^{20}0.005(20-x)dx=0.005(20x-\frac{x^2}{2})|_{x=10}^{x=20}=0.25.$

$P(X>18)=\int_{18}^{20}0.05(20-x)dx=0.05(20x-\frac{x^2}{2})|_{x=18}^{x=20}=0.01.$

$P(X>18|P(X>10)=\frac{P(X>18)}{P(X>10)}=\frac{0.01}{0.25}=0.04.$