Question

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 32% of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random. Let X denote the number among the four who have earthquake insurance.(a) Find the probability distribution of X.

Answer 1

Answer:

$P(X = 0) = 0.2138$

$P(X = 1) = 0.4025$

$P(X = 2) = 0.2841$

$P(X = 3) = 0.0891$

$P(X = 4) = 0.0105$

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.

32% of all homeowners are insured against earthquake damage.

This means that $p = 0.32$

Four homeowners are to be selected at random.

This means that $n = 4$

Find the probability distribution of X.

This is the probability of each outcome

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

$P(X = 0) = C_{4,0}.(0.32)^{0}.(0.68)^{4} = 0.2138$

$P(X = 1) = C_{4,1}.(0.32)^{1}.(0.68)^{3} = 0.4025$

$P(X = 2) = C_{4,2}.(0.32)^{2}.(0.68)^{2} = 0.2841$

$P(X = 3) = C_{4,3}.(0.32)^{3}.(0.68)^{1} = 0.0891$

$P(X = 4) = C_{4,4}.(0.32)^{4}.(0.68)^{0} = 0.0105$