Respuesta :
Answer:
a) 8.74% probability that none of the people are Independent
b) 99.63% probability that fewer than 7 are Independent.
c) 0.37% probability that more than 6 people are Independent
Step-by-step explanation:
For each voter, there are only two possible outcomes. Either they are independent, or they are not. The probability of a person being independent is not related to another person. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
15% of voters are Independent
This means that [tex]p = 0.15[/tex]
15 people
This means that [tex]n = 15[/tex]
a) What is the probability that none of the people are Independent?
This is [tex]P(X = 0)[/tex].
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{15,0}.(0.15)^{0}.(0.85)^{15} = 0.0874[/tex]
8.74% probability that none of the people are Independent
b) What is the probability that fewer than 7 are Independent?
This is [tex]P(X < 7) = P(X \leq 6)[/tex].
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{15,0}.(0.15)^{0}.(0.85)^{15} = 0.0874[/tex]
[tex]P(X = 1) = C_{15,1}.(0.15)^{1}.(0.85)^{14} = 0.2312[/tex]
[tex]P(X = 2) = C_{15,2}.(0.15)^{2}.(0.85)^{13} = 0.2856[/tex]
[tex]P(X = 3) = C_{15,3}.(0.15)^{3}.(0.85)^{12} = 0.2184[/tex]
[tex]P(X = 4) = C_{15,4}.(0.15)^{4}.(0.85)^{11} = 0.1156[/tex]
[tex]P(X = 5) = C_{15,5}.(0.15)^{5}.(0.85)^{10} = 0.0449[/tex]
[tex]P(X = 6) = C_{15,6}.(0.15)^{6}.(0.85)^{9} = 0.0132[/tex]
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0874 + 0.2312 + 0.2856 + 0.2184 + 0.1156 + 0.0449 + 0.0132 = 0.9963[/tex]
99.63% probability that fewer than 7 are Independent.
c) What is the probability that more than 6 people are Independent?
Either 6 or less people are independent, or more than 6 are. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 6) + P(X > 6) = 1[/tex]
From b), we have that [tex]P(X \leq 6) = 0.9963[/tex] and we want [tex]P(X > 6)[/tex]. So
[tex]P(X > 6) = 1 - P(X \leq 6) = 1 - 0.9963 = 0.0037[/tex]
0.37% probability that more than 6 people are Independent
