Find the probability distribution of the number of green balls drawn when 3 balls are drawn one by one without replacement from a bag containing 3 green and 5 white balls.

Answer

To find the probability distribution of the number of green balls drawn when 3 balls are drawn one by one without replacement from a bag containing 3 green and 5 white balls, we can follow these steps:

### Step 1: Identify the Total Number of Balls
We have a total of 3 green balls and 5 white balls in the bag. Therefore, the total number of balls is:
\[ \text{Total Balls} = 3 + 5 = 8 \]

### Step 2: Define the Random Variable
Let \( X \) be the random variable representing the number of green balls drawn. The possible values of \( X \) are \( 0, 1, 2, \) and \( 3 \).

### Step 3: Calculate the Probability for Each Value of \( X \)

#### Case 1: \( X = 0 \) (No green balls drawn)
This means all three balls drawn are white. The probability can be calculated as:
\[ P(X = 0) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{5 \times 4 \times 3}{8 \times 7 \times 6} = \frac{60}{336} = \frac{5}{28} \]

#### Case 2: \( X = 1 \) (One green ball drawn)
The possible sequences are \( GWW, WGW, WWG \). The probability for each sequence is:
1. \( GWW \): \( \frac{3}{8} \times \frac{5}{7} \times \frac{4}{6} \)
2. \( WGW \): \( \frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} \)
3. \( WWG \): \( \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \)

Calculating the probabilities:
\[ P(X = 1) = 3 \left( \frac{3}{8} \times \frac{5}{7} \times \frac{4}{6} \right) = 3 \left( \frac{60}{336} \right) = \frac{180}{336} = \frac{15}{28} \]

#### Case 3: \( X = 2 \) (Two green balls drawn)
The possible sequences are \( GGW, GWG, WGG \). The probability for each sequence is:
1. \( GGW \): \( \frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \)
2. \( GWG \): \( \frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} \)
3. \( WGG \): \( \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \)

Calculating the probabilities:
\[ P(X = 2) = 3 \left( \frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \right) = 3 \left( \frac{30}{336} \right) = \frac{90}{336} = \frac{15}{56} \]

#### Case 4: \( X = 3 \) (Three green balls drawn)
This means all three balls drawn are green. The probability is:
\[ P(X = 3) = \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{6}{336} = \frac{1}{56} \]

### Step 4: Compile the Probability Distribution
Now we can summarize the probabilities in a table:

| \( X \) | Probability \( P(X) \) |
|---------|--------------------------|
| 0 | \( \frac{5}{28} \) |
| 1 | \( \frac{15}{28} \) |
| 2 | \( \frac{15}{56} \) |
| 3 | \( \frac{1}{56} \) |

### Step 5: Normalize the Probabilities
To ensure the probabilities sum to 1, we can convert them to a common denominator (e.g., 56):
- \( P(X=0) = \frac{10}{56} \)
- \( P(X=1) = \frac{30}{56} \)
- \( P(X=2) = \frac{15}{56} \)
- \( P(X=3) = \frac{1}{56} \)

### Final Probability Distribution Table
| \( X \) | Probability \( P(X) \) |
|---------|--------------------------|
| 0 | \( \frac{10}{56} \) |
| 1 | \( \frac{30}{56} \) |
| 2 | \( \frac{15}{56} \) |
| 3 | \( \frac{1}{56} \) |