The loss due to a fire in a commercial building is modeled by a random variable [math]X[/math] with density function

[[math]] f(x) = \begin{cases} 0.005(20 − x), \, 0\lt x \lt 20 \\ 0, \, \textrm{otherwise.} \end{cases} [[/math]]

Given that a fire loss exceeds 8, calculate the probability that it exceeds 16.

• 1/25
• 1/9
• 1/8
• 1/3
• 3/7

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: B

Note that

[[math]] \begin{align*} \operatorname{P}[X \gt x] = \int_x^{20} 0.005(20-t) dt &= 0.005 (20t - \frac{1}{2} t^2 ) \Big |_{t=0}^{20} \\ &= 0.005(400 - 200 - 20x + \frac{1}{2}x^2) \\ &= 0.005(200-20x + \frac{1}{2} x^2) \end{align*} [[/math]]

where [math] 0 \lt x \lt 20 [/math]. Therefore

[[math]] \operatorname{P}[ X \gt 16 X \gt 8] = \frac{\operatorname{P}[ X \gt 16]}{\operatorname{P}[ X \gt 8]} = \frac{200 - 20(16) + \frac{1}{2}(16)^2}{200-20(8) + \frac{1}{2} (8)^2} = \frac{8}{72} = \frac{1}{9}. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.