Consider a bag containing five blue and five red balls. Construct a probability distribution table which summarises the sampling distribution of the sample proportion of blue balls when samples of size 3 are taken from the bag, without replacement.

To construct a probability distribution table for the sample proportion of blue balls when samples of size 3 are taken from the bag without replacement, we follow these steps:

### Step 1: Define the Situation
- The bag contains 5 blue and 5 red balls.
- Samples of size 3 are drawn without replacement.
- We are interested in the sample proportion of blue balls in these samples.

### Step 2: Determine Possible Outcomes
When drawing 3 balls from the bag, the number of blue balls (\( X \)) in the sample can be:
- 0 blue balls (all 3 are red)
- 1 blue ball (2 red, 1 blue)
- 2 blue balls (2 blue, 1 red)
- 3 blue balls (all 3 are blue)

The sample proportion of blue balls (\( \hat{p} \)) is:
\[
\hat{p} = \frac{X}{3}
\]
where \( X \) is the number of blue balls in the sample.

Thus, the possible sample proportions are:
\[
\hat{p} \in \left\{ 0, \frac{1}{3}, \frac{2}{3}, 1 \right\}
\]

### Step 3: Calculate Probabilities for Each Outcome

We use combinatorics to calculate the probability of each outcome.

- **Total number of ways to choose 3 balls from 10:**  
\[
\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\]

#### Case 1: \( X = 0 \) (No blue balls, all 3 red)
- Number of ways to choose 3 red balls from 5 red balls:  
\[
\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10
\]
- Probability:  
\[
P(X = 0) = \frac{10}{120} = 0.0833
\]

#### Case 2: \( X = 1 \) (1 blue ball, 2 red balls)
- Number of ways to choose 1 blue ball from 5 blue balls:  
\[
\binom{5}{1} = 5
\]
- Number of ways to choose 2 red balls from 5 red balls:  
\[
\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10
\]
- Total number of ways to choose 1 blue and 2 red balls:  
\[
5 \times 10 = 50
\]
- Probability:  
\[
P(X = 1) = \frac{50}{120} = 0.4167
\]

#### Case 3: \( X = 2 \) (2 blue balls, 1 red ball)
- Number of ways to choose 2 blue balls from 5 blue balls:  
\[
\binom{5}{2} = 10
\]
- Number of ways to choose 1 red ball from 5 red balls:  
\[
\binom{5}{1} = 5
\]
- Total number of ways to choose 2 blue and 1 red ball:  
\[
10 \times 5 = 50
\]
- Probability:  
\[
P(X = 2) = \frac{50}{120} = 0.4167
\]

#### Case 4: \( X = 3 \) (All 3 are blue balls)
- Number of ways to choose 3 blue balls from 5 blue balls:  
\[
\binom{5}{3} = 10
\]
- Probability:  
\[
P(X = 3) = \frac{10}{120} = 0.0833
\]

### Step 4: Create the Probability Distribution Table

The sample proportion \( \hat{p} \) and corresponding probabilities \( P(\hat{p}) \) are as follows:

| Sample Proportion (\( \hat{p} \)) | Probability \( P(\hat{p}) \) |
|:---------------------------------:|:----------------------------:|
| 0                                | 0.0833                       |
| \( \frac{1}{3} \)                 | 0.4167                       |
| \( \frac{2}{3} \)                 | 0.4167                       |
| 1                                | 0.0833                       |

This table summarizes the sampling distribution of the sample proportion of blue balls when samples of size 3 are taken from the bag without replacement.

---

Would you like further explanation on any of these calculations? Here are 5 related questions for deeper exploration:

1. How does the sampling distribution change if the sample size is increased to 4?
2. What is the expected value of the sample proportion \( \hat{p} \)?
3. How would the probabilities change if we sampled with replacement?
4. How does the variance of the sample proportion \( \hat{p} \) compare to other sample sizes?
5. Can you explain the concept of hypergeometric distribution and how it applies here?

**Tip:** When drawing samples without replacement, you are dealing with a hypergeometric distribution, unlike the binomial distribution, which assumes replacement.

Explore the sampling distribution of the sample proportion of blue balls when drawing 3 balls from a bag containing 5 blue and 5 red balls without replacement. This problem covers concepts in probability, combinatorics, and hypergeometric distribution, suitable for Grades 11-12.

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