Question

in the united states, voters who are neither democrat nor republican are called independent. it is believed that 11% of voters are independent. a survey asked 20 people to identify themselves as democrat, republican, or independent.
what is the probability that none of the people are independent? (enter your answer to 6 digits after the decimal place.)
probability =
what is the probability that fewer than 7 are independent?
probability =
what is the probability that more than 2 people are independent?
probability =
note: you can earn partial credit on this problem.

Explanation:

Step1: Identify the probability of non - independent

The probability that a voter is independent is $p = 0.11$, so the probability that a voter is not independent is $q=1 - p=1 - 0.11 = 0.89$. We use the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, $q$ is the probability of failure on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Calculate the probability that none are independent

For $n = 20$ and $k = 0$, we have $P(X = 0)=C(20,0)\times(0.11)^{0}\times(0.89)^{20}$. Since $C(20,0)=1$ and $(0.11)^{0}=1$, then $P(X = 0)=(0.89)^{20}\approx0.107427$.

Step3: Calculate the probability that fewer than 7 are independent

$P(X\lt7)=\sum_{k = 0}^{6}C(20,k)\times(0.11)^{k}\times(0.89)^{20 - k}$.
$C(20,0)\times(0.11)^{0}\times(0.89)^{20}+C(20,1)\times(0.11)^{1}\times(0.89)^{19}+C(20,2)\times(0.11)^{2}\times(0.89)^{18}+C(20,3)\times(0.11)^{3}\times(0.89)^{17}+C(20,4)\times(0.11)^{4}\times(0.89)^{16}+C(20,5)\times(0.11)^{5}\times(0.89)^{15}+C(20,6)\times(0.11)^{6}\times(0.89)^{14}$
$C(20,0)=1$, $C(20,1)=\frac{20!}{1!(20 - 1)!}=20$, $C(20,2)=\frac{20!}{2!(20 - 2)!}=190$, $C(20,3)=\frac{20!}{3!(20 - 3)!}=1140$, $C(20,4)=\frac{20!}{4!(20 - 4)!}=4845$, $C(20,5)=\frac{20!}{5!(20 - 5)!}=15504$, $C(20,6)=\frac{20!}{6!(20 - 6)!}=38760$
After calculation, $P(X\lt7)\approx0.995292$.

Step4: Calculate the probability that more than 2 are independent

$P(X\gt2)=1 - P(X\leq2)=1-(P(X = 0)+P(X = 1)+P(X = 2))$
$P(X = 0)=(0.89)^{20}$, $P(X = 1)=C(20,1)\times(0.11)^{1}\times(0.89)^{19}=20\times0.11\times(0.89)^{19}$, $P(X = 2)=C(20,2)\times(0.11)^{2}\times(0.89)^{18}=190\times0.0121\times(0.89)^{18}$
$P(X\gt2)\approx1-(0.107427 + 20\times0.11\times(0.89)^{19}+190\times0.0121\times(0.89)^{18})\approx0.558442$

Answer:

Probability that none are independent: $0.107427$
Probability that fewer than 7 are independent: $0.995292$
Probability that more than 2 are independent: $0.558442$