Q. 7.68 - A First Course in Probability

Question

The number of accidents that a person has in a given year is a Poisson random variable with mean $\lambda$̣  However, suppose that the value of $\lambda$ changes from person to person, being equal to $2$ for $60$ percent of the population and $3$ for the other $40$ percent. If a person is chosen at random, what is the probability that he will have

(a) $0$ accidents and,

(b) Exactly $3$ accidents in a certain year? What is the conditional probability that he will have $3$ accidents in a given year, given that he had no accidents the preceding year? 

Step-by-Step Solution

Answer

a. A number of $0$ accidents value are $P\{X=0\}=0.10112$.

b. The exact $3$accidents in a certain year value are $P\{X=3\}=0.1979$and $P\{X=3\mid$he had no accidents the preceding year$\}=0.189$. 

Step 1: Given Information (Part a)

A number of accidents that a person has in a given year is a Poisson random variable with mean$\lambda$.

Step 2: Explanation (Part a)

Let $X$ is a number of accidents that a person has in a given year is a Poisson random variable with mean $\lambda$.

$\lambda =2$ with $P\{\lambda =2\}=0.6$ and $\lambda =3$ 

With $P\{\lambda =3\}=0.4$

$P\{X=0\}=e^{-\lambda }$.

Step 3: Explanation (Part a)

Substitute

$P\{X=0\}=P\{X=0\mid \lambda =2\}+P\{X=0\mid \lambda =3\}$

$=0.6·e^{-2}+0.4·e^{-3}$

Add the value,

$=0.10112$.

Step 4: Final answer (Part a)

A number of $0$ accidents value found to be $P\{X=0\}=0.10112$.

Step 5: Given Information (Part b)

 Exactly $3$accidents in a certain year and given that he had no accidents the preceding year 

Step 6: Explanation (Part b)

Exactly$3$ accidents in a certain year

$P\{X=3\}=P\{X=3\mid \lambda =2\}+P\{X=3\mid \lambda =3\}$

Substitute, 

$=0.6\frac{2^3{e}^{-2}}{3!}+0.4\frac{3^{3e^{-3}}}{3!}$

$=0.6\frac{8e^{-2}}{6}+0.4\frac{27e^{-3}}{6}$

Simplify,

$=\frac{1}{6}(0.645+0.54)$

$=0.1979$.

Step 7: Explanation (Part b)

$P\{X=3\mid$he had no accidents the preceding year$\}$

$=\frac{P\{X=3,X=0\}}{P\{X=0\}}$

Substitute,

$=\frac{0.6\frac{2^3{e}^{-2}}{3!}{e}^{-2}+0.4\frac{3^{3e^{-3}}}{3!}{e}^{-3}}{0.10112}$

Divide,

$=\frac{0.019114265}{0.10112}$

$=0.189$

Step 8: Final answer (Part b)

The exact $3$ accidents in a certain year value are $P\{X=3\}=0.1979$ and $P\{X=31$he had no accidents the preceding year$\}=0.189$.