Question

An insurance company has discovered
chat only about 0.1 percent 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 5 of its clients are involved in such an accident next year?​

Answer 1

Step-by-step explanation:

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) = \begin{gathered}\left[\begin{array}{ccc}n\\\\p\end{array}\right]\end{gathered}

⎣

⎢

⎡

n

p

⎦

⎥

⎤

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

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