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.
