Some parts of California are particularl earthquake prone. Suppose in one metropolitan area, 40% of all homeowners are insured against earthquake damage. Four homeowners are selected at random. Let X denote the number among the four who have earthquake insurance. Assume that the event of each homeowner having or not having insurance is independent. Required:
a. Find the probability mass function 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?

Kindly check explanation

Step-by-step explanation:

Given that:

Percentage of insured home owners = 40% p = 0.4

Selecting 4 home owners (x) ;

x = 0, 1, 2, 3, 4

Using bernoulli;

nCx * p^x * (1 - p)^(n-x)

For X = 0

4C0 * 0.4^0 * 0.6^(4-0) = 0.1296

X = 1

4C1 * 0.4^1 * 0.6^(4-1) = 0.3456

X = 2

4C2 * 0.4^2 * 0.6^(4-2) = 0.3456

X = 3

4C3 * 0.4^3 * 0.6^(4-3) = 0.1536

X = 4

4C4 * 0.4^4 * 0.6^(4-4) = 0.0256

B.) The most likely value for X are 1 and 2 with probabilities of 0.3456 respectively.

C.) probability that atleast 2 of the 4 selected have earthquake insurance ;

P(X ≥ 2) = p(2) + p(3) + p(4)

P(X ≥ 2) = 0.3456 + 0.1536 + 0.0256

P(X ≥ 2) = 0.5248

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.