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

The probability that at least one out of 20 randomly chosen homeowners in California has earthquake insurance is approximately 0.999. The probability that four or more have earthquake insurance is approximately 0.867, calculated using the binomial probability formula.

Step-by-step explanation:

To answer the student's question, we use the concepts of probability, specifically the binomial probability formula. Considering that 3 out of 10 homeowners in California have earthquake insurance, the probability of a randomly chosen homeowner having earthquake insurance is 0.3, and the probability of not having it is 0.7.

Probability of at least one homeowner having earthquake insurance

The probability of at least one homeowner out of 20 having insurance is calculated by subtracting the probability of none having insurance from 1:

P(at least one) = 1 - P(none)
= 1 - (0.7)^20

Using a calculator:

P(at least one) = 1 - (0.7)^20
≈ 0.999

Probability of four or more homeowners having earthquake insurance

The probability of four or more homeowners having insurance is calculated by summing the probabilities from four to twenty using the binomial probability formula:

P(X ≥ 4) = Σ[n=4 to 20] (Combination(20,n) * (0.3)^n * (0.7)^(20-n))

Using a binomial probability calculator or table, we find:

P(X ≥ 4) ≈ 0.867

Answer 2

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) = (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 = (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