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−x), for 0 < x < 20 0, elsewhere Given that the fire loss exceeds 10, what is the probability that it exceeds 18

Answer 1

$P\{X>18|X>10\}=\frac{P\{X>18, X>10\}}{P\{X>10\}}\\ P\{X>18, X>10\}=P\{X>18\}\\ P\{X>18|X>10\}=\frac{P\{X>18\}}{P\{X>10\}}\\ F(x)=\int_{-\infty}^xf(t)dt.\\ f(x)\text{ is a pdf, } F(x)\text{ is a CDF}.\\ \text{Let } 0<x<20.\text{ We have:}\\ F(x)=\int_{0}^x 0.005(20-t)dt=-0.005\frac{x^2}{2}+0.1x.\\ F(x)=0, x<0.\\ F(x)=1, x>20.\\ P\{X>18\}=1-P\{X\leq 18\}=1-F(18)=\\ =1+0.005\frac{18^2}{2}-0.1\cdot 18=0.01.\\ P\{X>10\}=1-P\{X\leq 10\}=1-F(10)=\\ =1+0.005\frac{10^2}{2}-0.1\cdot 10=0.25.\\ P\{X>18|X>10\}=\frac{0.01}{0.25}=0.04\\ \text{Probability that the loss due to an earthquake}\\ \text{will exceed 18 equals 0.04 (given that the loss exceeds 10)}.$