Question

Suppose that 3 out of every 10 homeowners in the state of California has invested in earthquake insurance. Suppose 20 homeowners are randomly chosen to be interviewed.
(a) What is the probability that at least one had earthquake insurance? (Round your answer to three decimal places.)
(b) What is the probability that four or more have earthquake insurance? (Round your answer to three decimal places.)

Answer 1

Answer:

a) 0.999 = 99.9% probability that at least one had earthquake insurance

b) 0.892 = 89.2% probability that four or more have earthquake insurance

Step-by-step explanation:

For each homeowner, there are only two possible outcomes. Either they have invested in earthquake insurance, or they have not. The probability of a home owner having invested in earthquake insurance is independent from other homeowners. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Suppose that 3 out of every 10 homeowners in the state of California has invested in earthquake insurance.

This means that $p = \frac{3}{1-0} = 0.3$

Suppose 20 homeowners are randomly chosen to be interviewed.

This means that $n = 20$.

(a) What is the probability that at least one had earthquake insurance? (Round your answer to three decimal places.)

Either none had earthquake insurance, or at least one did. The sum of the probabilities of these events is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$. So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.3)^{0}.(0.7)^{20} = 0.001$

So

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.001 = 0.999$

0.999 = 99.9% probability that at least one had earthquake insurance

(b) What is the probability that four or more have earthquake insurance? (Round your answer to three decimal places.)

Either less than 4 had earthquake insurance, or at least four did. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$. So

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.3)^{0}.(0.7)^{20} = 0.001$

$P(X = 1) = C_{20,1}.(0.3)^{1}.(0.7)^{19} = 0.007$

$P(X = 2) = C_{20,2}.(0.3)^{2}.(0.7)^{18} = 0.028$

$P(X = 3) = C_{20,3}.(0.3)^{3}.(0.7)^{17} = 0.072$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.001 + 0.007 + 0.028 + 0.072 = 0.108$

So

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.108 = 0.892$

0.892 = 89.2% probability that four or more have earthquake insurance