In a certain county, $$ 30 \% $$ of the voters are Republicans. Assume each voters' political affiliation is independent of one another. Part a: If ten voters are selected at random, find the probability that no more than six of them will be Republicans. Part b: If 200 people vote at a specific polling location, what is the standard deviation of the number of voters who are Republican? $$ \begin{array}{l} ext { Part c: What is the probability that the first voter to be a Republican at a } \ ext { given election would occur after the } 15^{h} ext { person to enter the polling } \ ext { location? } \ \begin{array}{l} ext { Part d: On average, how many voters would we expect to survey } \ ext { before finding a Republican voter? }\end{array}\end{array} $$.

Question

In a certain county, $$ 30 \% $$ of the voters are Republicans. Assume each voters' political affiliation is independent of one another. Part a: If ten voters are selected at random, find the probability that no more than six of them will be Republicans. Part b: If 200 people vote at a specific polling location, what is the standard deviation of the number of voters who are Republican? $$ \begin{array}{l}	ext { Part c: What is the probability that the first voter to be a Republican at a } \ 	ext { given election would occur after the } 15^{h} 	ext { person to enter the polling } \ 	ext { location? } \ \begin{array}{l}	ext { Part d: On average, how many voters would we expect to survey } \ 	ext { before finding a Republican voter? }\end{array}\end{array} $$.

UpStudy AI · Accepted Answer

To solve the problems, we will use the binomial distribution and properties of the geometric distribution.

### Part a: Probability that no more than six of ten voters are Republicans

Let $$ X $$ be the number of Republicans among 10 voters. Since each voter is independent and the probability of a voter being a Republican is $$ p = 0.3 $$, $$ X $$ follows a binomial distribution:

$$
X \sim \text{Binomial}(n=10, p=0.3)
$$

We want to find $$ P(X \leq 6) $$. This can be calculated as:

$$
P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
$$

The probability mass function for a binomial distribution is given by:

$$
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
$$

Calculating each term:

- $$ P(X = 0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 1 \cdot 1 \cdot 0.7^{10} \approx 0.0282475 $$
- $$ P(X = 1) = \binom{10}{1} (0.3)^1 (0.7)^9 = 10 \cdot 0.3 \cdot 0.7^9 \approx 0.1210608 $$
- $$ P(X = 2) = \binom{10}{2} (0.3)^2 (0.7)^8 = 45 \cdot 0.09 \cdot 0.7^8 \approx 0.2334744 $$
- $$ P(X = 3) = \binom{10}{3} (0.3)^3 (0.7)^7 = 120 \cdot 0.027 \cdot 0.7^7 \approx 0.2612736 $$
- $$ P(X = 4) = \binom{10}{4} (0.3)^4 (0.7)^6 = 210 \cdot 0.0081 \cdot 0.7^6 \approx 0.2001209 $$
- $$ P(X = 5) = \binom{10}{5} (0.3)^5 (0.7)^5 = 252 \cdot 0.00243 \cdot 0.7^5 \approx 0.1029191 $$
- $$ P(X = 6) = \binom{10}{6} (0.3)^6 (0.7)^4 = 210 \cdot 0.000729 \cdot 0.7^4 \approx 0.0367569 $$

Now, summing these probabilities:

$$
P(X \leq 6) \approx 0.0282475 + 0.1210608 + 0.2334744 + 0.2612736 + 0.2001209 + 0.1029191 + 0.0367569 \approx 0.9838533
$$

Thus, the probability that no more than six of the ten voters will be Republicans is approximately **0.9839**.

### Part b: Standard deviation of the number of Republican voters in 200 people

Let $$ Y $$ be the number of Republicans among 200 voters. $$ Y $$ also follows a binomial distribution:

$$
Y \sim \text{Binomial}(n=200, p=0.3)
$$

The standard deviation $$ \sigma $$ of a binomial distribution is given by:

$$
\sigma = \sqrt{n p (1-p)}
$$

Substituting the values:

$$
\sigma = \sqrt{200 \cdot 0.3 \cdot 0.7} = \sqrt{200 \cdot 0.21} = \sqrt{42} \approx 6.48
$$

Thus, the standard deviation of the number of Republican voters in 200 people is approximately **6.48**.

### Part c: Probability that the first voter to be a Republican occurs after the 15th person

This scenario follows a geometric distribution. The probability that the first success (a Republican voter) occurs after $$ k $$ trials is given by:

$$
P(X > k) = (1-p)^k
$$

For $$ k = 15 $$ and $$ p = 0.3 $$:

$$
P(X > 15) = (0.7)^{15} \approx 0.004828
$$

Thus, the probability that the first voter to be a Republican occurs after the 15th person is approximately **0.0048**.

### Part d: Expected number of voters to survey before finding a Republican voter

The expected value $$ E(X) $$ for a geometric distribution is given by:

$$
E(X) = \frac{1}{p}
$$

Substituting $$ p = 0.3 $$:

$$
E(X) = \frac{1}{0.3} \approx 3.33
$$

Thus, on average, we would expect to survey approximately **3.33** voters before finding a Republican voter.
Part a: The probability that no more than six of the ten voters are Republicans is approximately 0.9839. Part b: The standard deviation of the number of Republican voters in 200 people is approximately 6.48. Part c: The probability that the first voter to be a Republican occurs after the 15th person is approximately 0.0048. Part d: On average, we would expect to survey approximately 3.33 voters before finding a Republican voter.

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