Respuesta :
Answer:
(a) 0.3225
(b) 0.9997
(c) 0.01799
Step-by-step explanation:
Since 9% of voters are independent, p=0.09 and n=12, the required probability is,
[tex]P(X=0)=\left[\begin{array}{c}12\\0\\\end{array}\right](0.09)^{0}(1-0.09)^{12-0}[/tex]=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)=[tex](\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})[/tex]
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-{[tex](\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})}[/tex]
=1-(0.322475+0.382718161+0.2081819+0.068631)
=1-0.982007=0.01799311
P(X>3)= 0.01799
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
p is the probability of a success on a single trial.
In this problem:
- 9% are independent, thus [tex]p = 0.09[/tex]
- 12 voters, thus [tex]n = 12[/tex]
Item a:
This probability is P(X = 0), thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225[/tex]
0.3225 = 32.25% probability that none of the people are Independent.
Item b:
This is:
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225[/tex]
[tex]P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827[/tex]
[tex]P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082[/tex]
[tex]P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686[/tex]
[tex]P(X = 4) = C_{12,4}.(0.09)^{4}.(0.91)^{8} = 0.0153[/tex]
[tex]P(X = 5) = C_{12,5}.(0.09)^{5}.(0.91)^{7} = 0.0024[/tex]
[tex]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[/tex]
0.9997 = 99.97% probability that fewer than 6 are Independent.
Item c:
This probability is:
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
From item b:
[tex]P(X \leq 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982[/tex]
[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.982 = 0.018[/tex]
0.018 = 1.8% probability that more than 3 people are Independent.
A similar problem is given at https://brainly.com/question/13836999
