Question

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?

Answer 1

Missing part of the question

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

Answer:

`P(X = x) = ^4C_x * 0.25^x * 0.75^(4-x)`

`Mean = 1`

`Pr = 0.2617`

Explanation:

Given

`n = 4`

`p = 25\%`

Solving (a): The probability distribution of x

We have:

`n = 4`

`p = 25\%`
`= 0.25`

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

`q = 1 - p`

`q = 1 - 0.25`
`= 0.75`

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

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

`P(X = x) = ^nC_x * 0.25^x * 0.75^(n-x)`

Substitute 4 for n

`P(X = x) = ^4C_x * 0.25^x * 0.75^(4-x)`

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

We have:

`n = 4` and

`p = 25\%`
`= 0.25`

`Mean = np`

`Mean = 4 * 0.25`

`Mean = 1`

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

This is calculated as

`Pr = P(x = 2) + P(x = 3) + P(x = 4)`

`P(X = x) = ^4C_x * 0.25^x * 0.75^(4-x)`

`P(X=2) = ^4C_2 * 0.25^2 * 0.75^(4-2)`

`P(X=2) = 6 * 0.25^2 * 0.75^2 = 0.2109375`

`P(X=3) = ^4C_3 * 0.25^3 * 0.75^(4-3)`

`P(X=3) = 4 * 0.25^3 * 0.75 = 0.046875`

`P(X=4) = ^4C_4 * 0.25^4 * 0.75^(4-4)`

`P(X=4) = 1 * 0.25^4 * 1= 0.00390625`

So:

`Pr = P(x = 2) + P(x = 3) + P(x = 4)`

`Pr = 0.2109375 + 0.046875 + 0.00390625`

`Pr = 0.26171875`

`Pr = 0.2617` --- approximated