Question

An insurance company found that only 0.01% of the population is involved in a certain type of accident each year. If its 1000 policy holders were randomly selected from the population, what is the probability that not more than two of its clients are involved in such an accident next year? (

Answer 1

Answer:

0.095021

Step-by-step explanation:

We take this to be a binomial distribution with parameters n and p.

Bin (n,p)

In our case :

n = 1000

p = 1/10000 = 0.0001

We want the probability that not more than 2 clients are involved in an accident.

P(x ≤ 2)

This equals to : P(x = 1) + P(x = 2)

P(x = x) = $\left[\begin{array}{ccc}n\\\\p\end{array}\right]$ pˣ (1 - p)ⁿ⁻ˣ

Now :

P(x = 1) = 1000C¹ × (0.0001) × (1 - 0.0001)⁹⁹⁹ = 0.0905

P(x = 2) = 1000C² × (0.0001)² × (1 - 0.0001)⁹⁹⁸ = 0.004521

P(x ≤2) = 0.0905 + 0.004521 = 0.095021