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 0 of those selected would do internet voting? (round to five decimal places as needed.)

Explanation:

Step1: Identify the 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 $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Determine the values of $n$, $k$, and $p$

Here, $n = 12$ (the number of selected voters), $k = 0$ (the number of voters who do internet - voting), and $p=0.35$ (the probability that a voter would do internet - voting). Then $1 - p = 1-0.35 = 0.65$.

Step3: Calculate the combination $C(n,k)$

$C(12,0)=\frac{12!}{0!(12 - 0)!}=\frac{12!}{12!}=1$.

Step4: Calculate the probability

$P(X = 0)=C(12,0)\times(0.35)^{0}\times(0.65)^{12}$. Since any non - zero number to the power of 0 is 1, $(0.35)^{0}=1$. Then $P(X = 0)=1\times1\times(0.65)^{12}$.
$(0.65)^{12}\approx0.00568$.

Answer:

$0.00568$