Answer to Question #239588 in Statistics and Probability for hema

Question

An insurance company insures 4,000 people against loss of both eyes in a car accident. Based on

previous data, the rates were computed on the assumptions that on the average 10 persons in

1,00,000 will have car accident each year that result in this type of injury. What is the probability

that more than 3 of the insured will collect on their policy in a given year?

Answer 1

From the given information, we can use Poisson distribution to find the probability that more than three of the insured will collect on their policy in it given year

$n=4000 \\ p = \frac{10}{100000}=0.0001 \\ mean = λ=np = 4000 \times 0.0001 = 0.4 \\ P(X=x)= \frac{e^{- λ} λ^x}{x!}$

X = more than three of the insured will collect on their policy in it given year.

$P(X≥3) = 1 -P(X<3) \\ = 1 -[P(X=0) +P(X=1) +P(X=2)] \\ = 1 -[\frac{e^{-0.4}0.4^0}{0!}+\frac{e^{-0.4}0.4^1}{1!}+\frac{e^{-0.4}0.4^2}{2!}] \\ = 1 -[0.6703+0.2681+0.054] \\ = 1 -0.9920 \\ = 0.0079$

Therefore, the probability that more than three of the insured will collect on their policy in it given year is 0.008.