Question

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 33% 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.B) What is the most likely value for X?
C) What is the probability that at least two of the four selected have earthquake insurance?

Answer 1

Answer:

(a) The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b) The most likely value for X is 1.32.

(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.

Step-by-step explanation:

The random variable X is defined as the number among the four homeowners  who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, p = 0.33.

The random sample of homeowners selected is, n = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable X follows a Binomial distribution with parameters n = 4 and p = 0.33.

The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected value of X as follows:

$E(X)=np$

         $=4\times 0.33\\=1.32$

Thus, the most likely value for X is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              $=1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015$

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.