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

Answer:

Following are the solution to the given points:

Step-by-step 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:

$P(X=0) = \frac{1}{(1-0.2)} \int_{0.2}^{1} e^{-\lambda} d\lambda \\\\P(X=0) = \frac{1}{0.8} \times [-e^{-1}-(-e^{-0.2})] = 0.56356$

For class 2:

$P(X=0) = \frac{1}{(2-0.4)} \int_{0.4}^{2} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{1.6} \times [-e^{-2}-(-e^{-0.4})] = 0.33437$

For class 3:

$P(X=0) = \frac{1}{(3-0.6)} \int_{0.6}^{3} e^{-\lambda} d\lambda\\\\P(X=0) = \frac{1}{2.4} \times [-e^{-3}-(-e^{-0.6})] = 0.20793$

The population is equally divided into three classes of drivers.

Hence, the Probability

$\to P(X=0) = \frac{1}{3} \times 0.56356+\frac{1}{3} \times 0.33437+\frac{1}{3} \times 0.20793=0.36862$