Question

based on a survey, 35% 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 12 likely voters are randomly selected. complete parts (a) through (c) below.
a. what is the probability that exactly 9 of those selected would do internet voting?
0.00476
(round to five decimal places as needed.)
b. if 9 of the selected voters would do internet voting, is 9 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 9 or more is
, which is not low.
b. yes, because the probability of 9 or more is
, which is not low.
c. yes, because the probability of 9 or more is 0.00560, which is low.
d. no, because the probability of 9 or more is
, which is low.
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, $n = 12$, $p=0.35$, and for part (a) $k = 9$.

Step2: Calculate $C(12,9)$

$C(12,9)=\frac{12!}{9!(12 - 9)!}=\frac{12!}{9!3!}=\frac{12\times11\times10}{3\times2\times1}=220$.

Step3: Calculate $P(X = 9)$

$P(X = 9)=C(12,9)\times(0.35)^{9}\times(1 - 0.35)^{12 - 9}=220\times(0.35)^{9}\times(0.65)^{3}$.
$P(X = 9)=220\times0.0000823543\times0.274625\approx0.00476$.

Step4: Calculate the probability of 9 or more for part (b)

$P(X\geq9)=P(X = 9)+P(X = 10)+P(X = 11)+P(X = 12)$.
$C(12,10)=\frac{12!}{10!(12 - 10)!}=66$, $C(12,11)=\frac{12!}{11!(12 - 11)!}=12$, $C(12,12)=\frac{12!}{12!(12 - 12)!}=1$.
$P(X = 10)=C(12,10)\times(0.35)^{10}\times(0.65)^{2}=66\times0.000028825\times0.4225\approx0.00081$.
$P(X = 11)=C(12,11)\times(0.35)^{11}\times(0.65)^{1}=12\times0.000010089\times0.65\approx0.00008$.
$P(X = 12)=C(12,12)\times(0.35)^{12}\times(0.65)^{0}=1\times0.00000353\times1\approx0.000004$.
$P(X\geq9)=0.00476+0.00081 + 0.00008+0.000004=0.00565\approx0.00560$.
Since $P(X\geq9)=0.00560$ which is low (usually, a probability less than 0.05 is considered low in significance testing), 9 is significantly high.

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

The probability of at least one is $P(X\geq1)=1 - P(X = 0)$.
$P(X = 0)=C(12,0)\times(0.35)^{0}\times(0.65)^{12}=1\times1\times(0.65)^{12}\approx0.00573$.
$P(X\geq1)=1 - 0.00573 = 0.99427\approx0.994$.

Answer:

a. $0.00476$
b. C. Yes, because the probability of 9 or more is $0.00560$, which is low
c. $0.994$