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.

Answer

To find the probability distribution of the number of defective bulbs in a sample of 4 bulbs drawn at random with replacement from a lot of 30 bulbs (which includes 6 defective bulbs), we can follow these steps:

### Step 1: Define the parameters
- Total number of bulbs (N) = 30
- Number of defective bulbs (D) = 6
- Number of non-defective bulbs (N - D) = 30 - 6 = 24
- Sample size (n) = 4

### Step 2: Calculate the probability of selecting a defective bulb
The probability of selecting a defective bulb (p) is given by:
\[ p = \frac{D}{N} = \frac{6}{30} = \frac{1}{5} \]

### Step 3: Calculate the probability of selecting a non-defective bulb
The probability of selecting a non-defective bulb (q) is:
\[ q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5} \]

### Step 4: Define the random variable
Let X be the random variable representing the number of defective bulbs in the sample of 4. X can take values from 0 to 4.

### Step 5: Use the binomial probability formula
Since we are sampling with replacement, the distribution of X follows a binomial distribution:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where:
- \( n \) = number of trials (4)
- \( k \) = number of successes (number of defective bulbs)
- \( p \) = probability of success (defective bulb)
- \( q \) = probability of failure (non-defective bulb)

### Step 6: Calculate 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 \( X = 1 \)**:
\[ P(X = 1) = \binom{4}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^3 = 4 \cdot \frac{1}{5} \cdot \left(\frac{64}{125}\right) = \frac{256}{625} \]

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 7: Summarize the probability distribution
The probability distribution of the number of defective bulbs (X) is:
- \( P(X = 0) = \frac{256}{625} \)
- \( P(X = 1) = \frac{256}{625} \)
- \( P(X = 2) = \frac{96}{625} \)
- \( P(X = 3) = \frac{16}{625} \)
- \( P(X = 4) = \frac{1}{625} \)