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:

$P(X = x) = ^4C_x * 0.25^x * 0.75^{4-x}$

$Mean = 1$

$Pr = 0.2617$

Step-by-step 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