Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let denote the number among the four who have earthquake insurance. 1. Find the probability distribution of X. [Hint: Let S denote a homeowner who has insurance and one who does not. Then one possible outcome is SFSS, with probability (.25)(.75)(.25)(.25) and associated X value 3. There are 15 other outcomes].
2. What is the most likely value for X?
3. What is the probability that at least two of the four selected have earthquake insurance?

Missing part of the question

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 25% of all homeowners.....................

Step-by-step explanation:

Given

Solving (a): The probability distribution of x

We have:

The probability of not having earthquake insurance (q) is:

If x has insurance, then n - x do not.

The distribution follows a binomial pattern. So, the probability distribution is:

Substitute 4 for n

Solving (b): The most likely value of x i.e. The mean

We have:

and

Solving (c): At least 2 of 4 selected have earthquake insurance.

This is calculated as

So:

--- approximated

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement