based on a survey, 35% of likely voters would be willing to vote by internet instead of the in - person…

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.\na. what is the probability that exactly 9 of those selected would do internet voting?\n(round to five decimal places as needed.)

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.\na. what is the probability that exactly 9 of those selected would do internet voting?\n(round to five decimal places as needed.)

Answer

Explanation:

Step1: Identify binomial 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 values of $n$, $k$, and $p$

Here, $n = 12$ (number of voters selected), $k = 9$ (number of voters willing to vote by internet), and $p=0.35$ (probability a voter is willing to vote by internet), $1 - p = 0.65$.

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

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

Step4: Calculate the probability

$P(X = 9)=C(12,9)\times(0.35)^{9}\times(0.65)^{12 - 9}=220\times(0.35)^{9}\times(0.65)^{3}$. $(0.35)^{9}\approx0.000075418$, $(0.65)^{3}=0.274625$. $P(X = 9)=220\times0.000075418\times0.274625\approx0.00447$.

Answer:

$0.00447$