Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 27% 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. [Hint: Let S denote a homeowner that has insurance and F one who does not. Then one possible outcome is SFSS, with probability (0.27)(0.73)(0.27)(0.27) and associated X value 3. There are 15 other outcomes.] (Round your answers to four decimal places.)
x 0 1 2 3 4
p(x) _ __
(c) What is the most likely value for X?
(d) What is the probability that at least two of the four selected have earthquake insurance? (Round your answer to four decimal places.)

For each homeowner, there are only two possible outcomes. Either they are insured against earthquake damage, or they are not. Homeowners are independent. This means that 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.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

27% of all homeowners are insured against earthquake damage.

This means that [tex]p = 0.27[/tex]

Four homeowners are to be selected at random.

This means that [tex]n = 4[/tex]

(a) Find the probability distribution of X.

This is the probability of each outcome. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{4,0}.(0.27)^{0}.(0.73)^{4} = 0.2840[/tex]

[tex]P(X = 1) = C_{4,1}.(0.27)^{1}.(0.73)^{3} = 0.4201[/tex]

[tex]P(X = 2) = C_{4,2}.(0.27)^{2}.(0.73)^{2} = 0.2331[/tex]

[tex]P(X = 3) = C_{4,3}.(0.27)^{3}.(0.73)^{1} = 0.0575[/tex]

[tex]P(X = 4) = C_{4,4}.(0.27)^{4}.(0.73)^{0} = 0.0053[/tex]

So the probability distribution is

P(X = 0) = 0.2840

P(X = 1) = 0.4201

P(X = 2) = 0.2331

P(X = 3) = 0.0575

P(X = 4) = 0.0053

(c) What is the most likely value for X?

P(X = 1) has the highest probability, so X = 1 is the most likely value.

(d) What is the probability that at least two of the four selected have earthquake insurance?

[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.2331 + 0.0575 + 0.0053 = 0.2959[/tex]

0.2959 = 29.59% probability that at least two of the four selected have earthquake insurance.