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

We need to compute

$$P(X>18|X>10)={P(X>18)\over P(X>10)}$$

$$P(X>10)=\displaystyle\int_{10}^{\infin}f(x)dx=\displaystyle\int_{10}^{20}0.005(20-x)dx=$$

$$=0.005\bigg[20x-{x^2\over 2}\bigg]\begin{matrix} 20 \\ 10 \end{matrix}=0.005(400-200-(200-50))=$$

$$=0.25$$

$$P(X>18)=\displaystyle\int_{18}^{\infin}f(x)dx=\displaystyle\int_{18}^{20}0.005(20-x)dx=$$

$$=0.005\bigg[20x-{x^2\over 2}\bigg]\begin{matrix} 20 \\ 18 \end{matrix}=0.005(400-200-(360-162))=$$

$$=0.01$$

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