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