From a lot of 30 bulbs, which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. Also find the mean and variance of the distribution.

Answer

To solve the problem, we need to find the probability distribution of the number of defective bulbs when drawing 4 bulbs from a lot of 30 bulbs (which includes 6 defectives) with replacement. We will also calculate the mean and variance of this distribution.

### Step 1: Identify the parameters
- Total bulbs = 30
- Defective bulbs = 6
- Non-defective bulbs = 30 - 6 = 24
- Probability of selecting a defective bulb (p) = Number of defective bulbs / Total bulbs = 6 / 30 = 1/5
- Probability of selecting a non-defective bulb (q) = 1 - p = 1 - 1/5 = 4/5
- Number of draws (n) = 4

### Step 2: Define the random variable
Let \( X \) be the random variable representing the number of defective bulbs drawn in the sample of 4 bulbs.

### Step 3: Use the Binomial distribution
Since we are drawing with replacement, the situation can be modeled using the Binomial distribution:
\[ P(X = x) = \binom{n}{x} p^x q^{n-x} \]
Where:
- \( n = 4 \) (number of trials)
- \( p = \frac{1}{5} \) (probability of success)
- \( q = \frac{4}{5} \) (probability of failure)
- \( x \) = number of defective bulbs drawn (can take values 0, 1, 2, 3, or 4)

### Step 4: Calculate the probabilities for each value of x
1. **For \( X = 0 \)**:
\[ P(X = 0) = \binom{4}{0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^4 = 1 \cdot 1 \cdot \left(\frac{256}{625}\right) = \frac{256}{625} \]

2. **For \( X = 1 \)**:
\[ P(X = 1) = \binom{4}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^3 = 4 \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{64}{125}\right) = \frac{256}{625} \]

3. **For \( X = 2 \)**:
\[ P(X = 2) = \binom{4}{2} \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^2 = 6 \cdot \left(\frac{1}{25}\right) \cdot \left(\frac{16}{25}\right) = \frac{96}{625} \]

4. **For \( X = 3 \)**:
\[ P(X = 3) = \binom{4}{3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^1 = 4 \cdot \left(\frac{1}{125}\right) \cdot \left(\frac{4}{5}\right) = \frac{16}{625} \]

5. **For \( X = 4 \)**:
\[ P(X = 4) = \binom{4}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^0 = 1 \cdot \left(\frac{1}{625}\right) \cdot 1 = \frac{1}{625} \]

### Step 5: Summarize the probability distribution
The probability distribution of \( X \) is:
- \( P(X = 0) = \frac{256}{625} \)
- \( P(X = 1) = \frac{256}{625} \)
- \( P(X = 2) = \frac{96}{625} \)
- \( P(X = 3) = \frac{16}{625} \)
- \( P(X = 4) = \frac{1}{625} \)

### Step 6: Calculate the mean and variance
The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p = 4 \cdot \frac{1}{5} = \frac{4}{5} = 0.8 \]

The variance \( \sigma^2 \) of a binomial distribution is given by:
\[ \sigma^2 = n \cdot p \cdot q = 4 \cdot \frac{1}{5} \cdot \frac{4}{5} = \frac{16}{25} = 0.64 \]

### Final Result
- Probability Distribution:
- \( P(X = 0) = \frac{256}{625} \)
- \( P(X = 1) = \frac{256}{625} \)
- \( P(X = 2) = \frac{96}{625} \)
- \( P(X = 3) = \frac{16}{625} \)
- \( P(X = 4) = \frac{1}{625} \)

- Mean \( \mu = 0.8 \)
- Variance \( \sigma^2 = 0.64 \)