Question

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.

Answer 1

Following are the solution to the given points:

Explanation:

As a result, Poisson for each driver seems to be the number of accidents.

Let X be the random vector indicating accident frequency.

Let,
`\lambda=`Expected accident frequency

`P(X=0) = e^(-\lambda)`

For class 1:

`P(X=0) = (1)/((1-0.2)) \int_(0.2)^(1) e^(-\lambda) d\lambda \\\\P(X=0) = (1)/(0.8) * [-e^(-1)-(-e^(-0.2))] = 0.56356`

For class 2:

`P(X=0) = (1)/((2-0.4)) \int_(0.4)^(2) e^(-\lambda) d\lambda\\\\P(X=0) = (1)/(1.6) * [-e^(-2)-(-e^(-0.4))] = 0.33437`

For class 3:

`P(X=0) = (1)/((3-0.6)) \int_(0.6)^(3) e^(-\lambda) d\lambda\\\\P(X=0) = (1)/(2.4) * [-e^(-3)-(-e^(-0.6))] = 0.20793`

The population is equally divided into three classes of drivers.

Hence, the Probability

`\to P(X=0) = (1)/(3) * 0.56356+(1)/(3) * 0.33437+(1)/(3) * 0.20793=0.36862`