Question

based on a survey, 37% of likely voters would be willing to vote by internet instead of the in - person traditional method of voting. for each of the following, assume that 15 likely voters are randomly selected. complete parts (a) through (c) below.
a. what is the probability that exactly 12 of those selected would do internet voting?
0.00075
(round to five decimal places as needed.)
b. if 12 of the selected voters would do internet voting, is 12 significantly high? why or why not?
select the correct choice below and fill in the answer box within your choice.
(round to five decimal places as needed.)
a. no, because the probability of 12 or more is

b. yes, because the probability of 12 or more is 0.00086, which is low
c. no, because the probability of 12 or more is

d. yes, because the probability of 12 or more is

c. find the probability that at least one of the selected likely voters would do internet voting

(round to three decimal places as needed.)

Explanation:

Step1: Identify binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 15$, $k = 12$, $p=0.37$, and $1 - p = 0.63$.

Step2: Calculate the combination $C(n,k)$

$C(15,12)=\frac{15!}{12!(15 - 12)!}=\frac{15!}{12!3!}=\frac{15\times14\times13}{3\times2\times1}=455$.

Step3: Calculate the binomial probability for part (a)

$P(X = 12)=C(15,12)\times(0.37)^{12}\times(0.63)^{3}$
$P(X = 12)=455\times(0.37)^{12}\times(0.63)^{3}\approx455\times1.39\times10^{- 6}\times0.250=0.00075$ (rounded to five decimal places).

Step4: Calculate the probability of 12 or more for part (b)

$P(X\geq12)=P(X = 12)+P(X = 13)+P(X = 14)+P(X = 15)$
$P(X = 13)=C(15,13)\times(0.37)^{13}\times(0.63)^{2}=\frac{15!}{13!2!}\times(0.37)^{13}\times(0.63)^{2}=105\times4.14\times10^{-7}\times0.397\approx0.00002$
$P(X = 14)=C(15,14)\times(0.37)^{14}\times(0.63)^{1}=15\times1.53\times10^{-7}\times0.63\approx0.000001$
$P(X = 15)=C(15,15)\times(0.37)^{15}\times(0.63)^{0}=(0.37)^{15}\approx4.49\times10^{-8}$
$P(X\geq12)=0.00075 + 0.00002+0.000001 + 4.49\times10^{-8}\approx0.00086$
Since the probability of 12 or more is low (less than 0.05), 12 is significantly high.

Step5: Calculate the probability of at least one for part (c)

The probability of at least one is $P(X\geq1)=1 - P(X = 0)$
$P(X = 0)=C(15,0)\times(0.37)^{0}\times(0.63)^{15}=1\times1\times(0.63)^{15}\approx0.0016$.
$P(X\geq1)=1-(0.63)^{15}\approx1 - 0.0016 = 0.998$ (rounded to three decimal places).

Answer:

a. $0.00075$
b. B. Yes, because the probability of 12 or more is $0.00086$, which is low
c. $0.998$