Find the probability distribution of the number of white balls drawn in a random of 3 balls without replacement from a bag of 4 white and 6 red balls. Also find the mean and variance of the distribution.

Answer

To solve the problem of finding the probability distribution of the number of white balls drawn from a bag containing 4 white and 6 red balls, we follow these steps:

### Step 1: Identify the total number of balls and the possible outcomes
We have a total of 10 balls (4 white and 6 red). We are drawing 3 balls without replacement. The possible number of white balls (X) that can be drawn can be 0, 1, 2, or 3.

### Step 2: Calculate the probabilities for each outcome

**Case 1: X = 0 (0 white balls drawn)**
To find the probability of drawing 0 white balls, we draw all 3 balls from the 6 red balls:
\[ P(X = 0) = \frac{\text{Number of ways to choose 0 white and 3 red}}{\text{Total ways to choose 3 balls from 10}} \]
\[ = \frac{\binom{4}{0} \cdot \binom{6}{3}}{\binom{10}{3}} = \frac{1 \cdot 20}{120} = \frac{20}{120} = \frac{1}{6} \]

**Case 2: X = 1 (1 white ball drawn)**
To find the probability of drawing 1 white ball, we choose 1 from the 4 white balls and 2 from the 6 red balls:
\[ P(X = 1) = \frac{\binom{4}{1} \cdot \binom{6}{2}}{\binom{10}{3}} = \frac{4 \cdot 15}{120} = \frac{60}{120} = \frac{1}{2} \]

**Case 3: X = 2 (2 white balls drawn)**
To find the probability of drawing 2 white balls, we choose 2 from the 4 white balls and 1 from the 6 red balls:
\[ P(X = 2) = \frac{\binom{4}{2} \cdot \binom{6}{1}}{\binom{10}{3}} = \frac{6 \cdot 6}{120} = \frac{36}{120} = \frac{3}{10} \]

**Case 4: X = 3 (3 white balls drawn)**
To find the probability of drawing 3 white balls, we choose all 3 from the 4 white balls:
\[ P(X = 3) = \frac{\binom{4}{3} \cdot \binom{6}{0}}{\binom{10}{3}} = \frac{4 \cdot 1}{120} = \frac{4}{120} = \frac{1}{30} \]

### Step 3: Summarize the probability distribution
The probability distribution can be summarized as follows:

| X (Number of White Balls) | P(X) |
|---------------------------|-----------|
| 0 | 1/6 |
| 1 | 1/2 |
| 2 | 3/10 |
| 3 | 1/30 |

### Step 4: Calculate the mean (expected value) of the distribution
The mean (μ) is calculated as:
\[ \mu = \sum (x \cdot P(X)) \]
\[ = 0 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{30} \]
\[ = 0 + \frac{1}{2} + \frac{6}{10} + \frac{3}{30} \]
\[ = \frac{1}{2} + \frac{3}{5} + \frac{1}{10} \]
Finding a common denominator (10):
\[ = \frac{5}{10} + \frac{6}{10} + \frac{1}{10} = \frac{12}{10} = \frac{6}{5} \]

### Step 5: Calculate the variance of the distribution
The variance (σ²) is calculated as:
\[ \sigma^2 = \sum (x^2 \cdot P(X)) - \mu^2 \]
Calculating \( \sum (x^2 \cdot P(X)) \):
\[ = 0^2 \cdot \frac{1}{6} + 1^2 \cdot \frac{1}{2} + 2^2 \cdot \frac{3}{10} + 3^2 \cdot \frac{1}{30} \]
\[ = 0 + \frac{1}{2} + \frac{12}{10} + \frac{9}{30} \]
Finding a common denominator (30):
\[ = \frac{15}{30} + \frac{36}{30} + \frac{9}{30} = \frac{60}{30} = 2 \]
Now, calculate variance:
\[ \sigma^2 = 2 - \left(\frac{6}{5}\right)^2 = 2 - \frac{36}{25} = \frac{50}{25} - \frac{36}{25} = \frac{14}{25} \]

### Final Results
- Probability Distribution:
- P(X=0) = 1/6
- P(X=1) = 1/2
- P(X=2) = 3/10
- P(X=3) = 1/30
- Mean (μ) = 6/5
- Variance (σ²) = 14/25