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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?

Respuesta :

MrRoyal

Missing part of the question

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

Answer:

[tex]P(X = x) = ^4C_x * 0.25^x * 0.75^{4-x}[/tex]

[tex]Mean = 1[/tex]

[tex]Pr = 0.2617[/tex]

Step-by-step explanation:

Given

[tex]n = 4[/tex]

[tex]p = 25\%[/tex]

Solving (a): The probability distribution of x

We have:

[tex]n = 4[/tex]

[tex]p = 25\%[/tex] [tex]= 0.25[/tex]

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

[tex]q = 1 - p[/tex]

[tex]q = 1 - 0.25[/tex] [tex]= 0.75[/tex]

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

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

[tex]P(X = x) = ^nC_x * 0.25^x * 0.75^{n-x}[/tex]

Substitute 4 for n

[tex]P(X = x) = ^4C_x * 0.25^x * 0.75^{4-x}[/tex]

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

We have:

[tex]n = 4[/tex] and

[tex]p = 25\%[/tex] [tex]= 0.25[/tex]

[tex]Mean = np[/tex]

[tex]Mean = 4 * 0.25[/tex]

[tex]Mean = 1[/tex]

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

This is calculated as

[tex]Pr = P(x = 2) + P(x = 3) + P(x = 4)[/tex]

[tex]P(X = x) = ^4C_x * 0.25^x * 0.75^{4-x}[/tex]

[tex]P(X=2) = ^4C_2 * 0.25^2 * 0.75^{4-2}[/tex]

[tex]P(X=2) = 6 * 0.25^2 * 0.75^2 = 0.2109375[/tex]

[tex]P(X=3) = ^4C_3 * 0.25^3 * 0.75^{4-3}[/tex]

[tex]P(X=3) = 4 * 0.25^3 * 0.75 = 0.046875[/tex]

[tex]P(X=4) = ^4C_4 * 0.25^4 * 0.75^{4-4}[/tex]

[tex]P(X=4) = 1 * 0.25^4 * 1= 0.00390625[/tex]

So:

[tex]Pr = P(x = 2) + P(x = 3) + P(x = 4)[/tex]

[tex]Pr = 0.2109375 + 0.046875 + 0.00390625[/tex]

[tex]Pr = 0.26171875[/tex]

[tex]Pr = 0.2617[/tex] --- approximated

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