Question

Suppose that in one metropolitan area, 25% of all home- owners are insured against earthquake damage. Four home-owners are to be selected at random; let X denote the number among the four who have earthquake insurance. a. Find the probability distribution of X. [Hint: Let S denote a homeowner who has insurance and F 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.]

Answer 1

Answer:

Step-by-step explanation:

Given that,

25% home owners are insecure of earthquake problem

If we select 4 home owners at random

let X denote the number among the four who have earthquake insurance

Let find the probability distribution of X

Let S- denotes a home owner who has insurance

Let F denotes a home owner who does not have insurance.

Then, P(S) =25% = 0.25

P(F) = 1 — P(S) = 1 —0.25

P(F) = 0.75

The possible outcomes of X is

X(0) = If no person has insurance

X(1) = If only 1 person has insurance

X(2) = If only 2 persons has insurance

X(3) = If only 3 persons has insurance

X(4) = If only 4 persons has insurance.

The cardinality of the sample space is n(C) = 2ⁿ = 2⁴ => 16

So, the sample space is given as

{FFFF, FFFS, FFSF, FFSS, FSFF, FSFS, FSSF, FSSS, SFFF, SFFS, SFSF, SFSS, SSFF, SSFS, SSSF, SSSS}

For X=0, the possible is {FFFF} i.e. no insurance, the one without insurance.

P(X) = 0.25×0.25×0.25×0.25

P(X) = 0.25⁴

P(X) = 0.00390625

For X=1, the possible outcome are

FFFS, FFSF, FSFF, SFFF

P(X=1) = 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75

P(X=1) = 0.046875

For X=2 the possible outcomes are

FFSS, FSFS, FSSF, SFFS, SFSF, SSFF,

P(X=2)=0.25²•0.75²+0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²

P(X=2) = 0.2109375

For X=3 the possible outcomes are

FSSS, SFSS, SSFS, SSSF

P(X=3) = 0.25•0.75³+0.25•0.75³+ 0.25•0.75³+0.25•0.75³

P(X=3) = 0.421875

X=4 the possible outcomes are

SSSS

P(X=4) = 0.75×0.75×0.75×0.75

P(X=4) = 0.31640625.

Or

Using normal distribution

P(X=k) = ⁿCk • 0.25^k • 0.75^(4-k)

So,

P(X=0) = 4C0 • 0.25^0 • 0.75^4

P(X=0) = 0.31640625

P(X=1) = 4C1 • 0.25^1 • 0.75^3

P(X=1) = 0.421875

P(X=2) = 4C2 • 0.25^2 • 0.75^2

P(X=2) = 0.2109375

P(X=3) = 4C3 • 0.25^3 • 0.75^1

P(X=3) = 0.046875

P(X=4) = 4C4 • 0.25^4 • 0.75^0

P(X=4) = 0.00390625.