Mathematics, 24.10.2019 08:43 BOYAWESOME0202

In the united states, voters who are neither democrat nor republican are called independent. it is believed that 9% of voters are independent. a survey asked 12 people to identify themselves as democrat, republican, or independent.

a. what is the probability that none of the people are independent?

b. what is the probability that fewer than 6 are independent?

c. what is the probability that more than 3 people are independent?

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Answer from: kaymillsaps

Using the binomial distribution, it is found that:

a) 0.3225 = 32.25% probability that none of the people are Independent.

b) 0.9997 = 99.97% probability that fewer than 6 are Independent.

c) 0.018 = 1.8% probability that more than 3 people are Independent.

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For each voter, there are only two possible outcomes. Either they are independent, or they are not. The probability of a voter being independent is independent of any other voter, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

p is the probability of a success on a single trial.

In this problem:

9% are independent, thus p = 0.09

12 voters, thus n = 12

Item a:

This probability is P(X = 0), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225$

0.3225 = 32.25% probability that none of the people are Independent.

Item b:

This is:

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225$

$P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827$

$P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082$

$P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686$

$P(X = 4) = C_{12,4}.(0.09)^{4}.(0.91)^{8} = 0.0153$

$P(X = 5) = C_{12,5}.(0.09)^{5}.(0.91)^{7} = 0.0024$

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.3225 + 0.3827 + 0.2082 + 0.0686 + 0.0153 + 0.0024 = 0.9997

0.9997 = 99.97% probability that fewer than 6 are Independent.

Item c:

This probability is:

$P(X 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

From item b:

$P(X \leq 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982$

$P(X 3) = 1 - P(X \leq 3) = 1 - 0.982 = 0.018$

0.018 = 1.8% probability that more than 3 people are Independent.

A similar problem is given at link

Answer from: skaterwolf1317

(a) 0.3225

(b) 0.9997

Step-by-step explanation:

Since 9% of voters are independent, p=0.09 and n=12, the required probability is,

$P(X=0)=\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{0}(1-0.09)^{12-0}$ =0.322475487

Therefore, p=0.3225

(b)

Probability that fewer than 6 are independent voters

P(X<6)=[P(X=0)+ P(X=1)+ P(X=2)+ P(X=3)+ P(X=4)+ P(X=5)]

P(X<6)= $(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{0}(1-0.09)^{12-0})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{1}(1-0.09)^{12-1})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{2}(1-0.09)^{12-2})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{3}(1-0.09)^{12-3})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{4}(1-0.09)^{12-4})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{5}(1-0.09)^{12-5})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{6}(1-0.09)^{12-6})$

P(X<6)=0.322475+0.382718161+0.2081819+0.068631+0.01527237+0.00241673= 0.999696

P(X<6)=0.9997

Therefore, probability that fewer than six people are independent is 0.9997

(c)

Probability that more than 3 people are independent voters

P(X>3)=1-P(X≤3)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)]

=1-{ $(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{0}(1-0.09)^{12-0})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{1}(1-0.09)^{12-1})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{2}(1-0.09)^{12-2})+(\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{3}(1-0.09)^{12-3})}$

=1-(0.322475+0.382718161+0.2081819+0.068631)

=1-0.982007=0.01799311

P(X>3)= 0.01799

Answer from: Quest

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step-by-step explanation:

Answer from: Quest

step-by-step explanation:

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