Mathematics, 20.07.2021 02:30 GhostFace18595
A population is equally divided into three class of drivers. The number of accidents per individual driver is Poisson for all drivers. For a driver of Class I, the expected number of accidents is uniformly distributed over [0.2, 1.0]. For a driver of Class II, the expected number of accidents is uniformly distributed over [0.4, 2.0]. For a driver of Class III, the expected number of accidents is uniformly distributed over [0.6, 3.0]. For driver randomly selected from this population, determine the probability of zero accidents.
Answers: 2
Expected accident frequency
![P(X=0) = \frac{1}{(1-0.2)} \int_{0.2}^{1} e^{-\lambda} d\lambda \\\\P(X=0) = \frac{1}{0.8} \times [-e^{-1}-(-e^{-0.2})] = 0.56356](/tpl/images/1396/5943/b34f1.png)
![P(X=0) = \frac{1}{(2-0.4)} \int_{0.4}^{2} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{1.6} \times [-e^{-2}-(-e^{-0.4})] = 0.33437](/tpl/images/1396/5943/8aa64.png)
![P(X=0) = \frac{1}{(3-0.6)} \int_{0.6}^{3} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{2.4} \times [-e^{-3}-(-e^{-0.6})] = 0.20793](/tpl/images/1396/5943/be887.png)
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A population is equally divided into three class of drivers. The number of accidents per individual...
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