An industry representative claims that 32% of all satellite dish owners subscribe to at least one premium movie channel. In an attempt to justify this claim, the representative will poll a randomly selected sample of dish owners.
 

**Round answers to four decimal places.**
 
**a)** Suppose that the representative's claim is true, and suppose that a sample of 4 dish owners is randomly selected. Calculate the following probabilities:
 
1. The probability that none of the dish owners in the sample subscribe to at least one premium movie channel.              


2. The probability that more than two dish owners in the sample subscribe to at least one premium movie channel. 
 
 
**b)** Suppose the representative's claim is true, and suppose that a sample of 20 dish owners is randomly selected. What is the probability that: 
 
1. At most nine dish owners in the sample subscribe to at least one premium movie channel? 


2. More than 11 dish owners in the sample subscribe to at least one premium movie channel? 
 

3. Fewer than five dish owners in the sample subscribe to at least one premium movie channel?   

We can approach these questions using the **binomial distribution**. In the binomial distribution, we have:

- \( n \) = number of trials (number of dish owners sampled)
- \( p \) = probability of success (dish owner subscribes to at least one premium movie channel)
- \( 1 - p \) = probability of failure (dish owner does not subscribe)

The general formula for the binomial probability is:

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

Where:
- \( X \) is the number of successes (dish owners subscribing to a premium movie channel),
- \( \binom{n}{k} \) is the binomial coefficient.

For this problem, the probability of success is \( p = 0.32 \), as 32% of dish owners are expected to subscribe.

---

### Part (a) - Sample size = 4
For this part, \( n = 4 \) and \( p = 0.32 \).

#### 1. The probability that none of the dish owners in the sample subscribe to at least one premium movie channel:
This means \( k = 0 \) successes (no one subscribes). Using the binomial formula:

\[
P(X = 0) = \binom{4}{0} (0.32)^0 (0.68)^4 = 1 \cdot 1 \cdot (0.68)^4
\]

Let's compute this value.

#### 2. The probability that more than two dish owners in the sample subscribe to at least one premium movie channel:
This means we need to calculate \( P(X > 2) \). This can be written as:

\[
P(X > 2) = P(X = 3) + P(X = 4)
\]

We will calculate both probabilities and add them.

---

### Part (b) - Sample size = 20
For this part, \( n = 20 \) and \( p = 0.32 \).

#### 1. The probability that at most nine dish owners in the sample subscribe:
This is \( P(X \leq 9) \), the cumulative probability from 0 to 9.

\[
P(X \leq 9) = \sum_{k=0}^{9} P(X = k)
\]

#### 2. The probability that more than 11 dish owners in the sample subscribe:
This is \( P(X > 11) \), or:

\[
P(X > 11) = 1 - P(X \leq 11)
\]

#### 3. The probability that fewer than five dish owners in the sample subscribe:
This is \( P(X < 5) \), or:

\[
P(X < 5) = P(X \leq 4)
\]

I'll now compute these probabilities.Here are the results for the given probabilities:

### Part (a)
1. The probability that none of the dish owners in the sample subscribe to at least one premium movie channel is approximately **0.2138**.
   
2. The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is approximately **0.0996**.

### Part (b)
1. The probability that at most nine dish owners in the sample subscribe to at least one premium movie channel is approximately **0.9281**.

2. The probability that more than 11 dish owners in the sample subscribe to at least one premium movie channel is approximately **0.0091**.

3. The probability that fewer than five dish owners in the sample subscribe to at least one premium movie channel is approximately **0.1827**.

---

Do you want further clarification or details on any of these results?

### Related Questions:
1. How does increasing the sample size affect the probabilities in a binomial distribution?
2. What is the mean and variance of the number of dish owners who subscribe to a premium channel in a sample of 20?
3. What is the probability that exactly 5 dish owners out of 20 subscribe to at least one premium movie channel?
4. How would the probabilities change if the probability of success increased to 50%?
5. How do these probabilities relate to a normal approximation for larger sample sizes?

### Tip:
For large sample sizes, the binomial distribution can be approximated using the normal distribution, especially when \( n \) is large and \( p \) is not too close to 0 or 1.

An industry representative claims that 32% of all satellite dish owners subscribe to at least one premium movie channel. In an attempt to justify this claim, the representative will poll a randomly selected sample of dish owners.

**Round answers to four decimal places.**

a) Suppose that the representative's claim is true, and suppose that a sample of 4 dish owners is randomly selected. Calculate the following probabilities:

1. The probability that none of the dish owners in the sample subscribe to at least one premium movie channel.

2. The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

b) Suppose the representative's claim is true, and suppose that a sample of 20 dish owners is randomly selected. What is the probability that:

1. At most nine dish owners in the sample subscribe to at least one premium movie channel?

2. More than 11 dish owners in the sample subscribe to at least one premium movie channel?

3. Fewer than five dish owners in the sample subscribe to at least one premium movie channel?

Explore the binomial distribution to calculate probabilities related to satellite dish owners subscribing to premium movie channels. Includes detailed step-by-step solutions for different sample sizes and success probabilities. Suitable for high school students in Grades 10-12.

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