It is given that 10% of the electric bulbs manufactured by a company are defective. In a sample of 20 bulbs, find the probability that more than 2 are defective.

Answer

To solve the problem of finding the probability that more than 2 out of 20 electric bulbs are defective, we can use the binomial probability formula. Here’s a step-by-step solution:

### Step 1: Identify the parameters
We know that:
- The probability of a bulb being defective (p) = 10% = 0.1
- The probability of a bulb being non-defective (q) = 1 - p = 1 - 0.1 = 0.9
- The number of trials (n) = 20 (the total number of bulbs)

### Step 2: Define the event
We want to find the probability that more than 2 bulbs are defective. This can be expressed mathematically as:
\[ P(X > 2) \]
where \( X \) is the random variable representing the number of defective bulbs.

### Step 3: Use the complement rule
Instead of calculating \( P(X > 2) \) directly, we can use the complement rule:
\[ P(X > 2) = 1 - P(X \leq 2) \]
This means we need to calculate the probability that 2 or fewer bulbs are defective.

### Step 4: Calculate \( P(X \leq 2) \)
We can find \( P(X \leq 2) \) by calculating \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

#### Calculate \( P(X = 0) \):
\[ P(X = 0) = \binom{20}{0} (0.1)^0 (0.9)^{20} = 1 \times 1 \times (0.9)^{20} \]
\[ P(X = 0) = (0.9)^{20} \approx 0.1216 \]

#### Calculate \( P(X = 1) \):
\[ P(X = 1) = \binom{20}{1} (0.1)^1 (0.9)^{19} = 20 \times (0.1) \times (0.9)^{19} \]
\[ P(X = 1) = 20 \times 0.1 \times (0.9)^{19} \approx 20 \times 0.1 \times 0.1351 \approx 0.2702 \]

#### Calculate \( P(X = 2) \):
\[ P(X = 2) = \binom{20}{2} (0.1)^2 (0.9)^{18} = \frac{20 \times 19}{2} \times (0.1)^2 \times (0.9)^{18} \]
\[ P(X = 2) = 190 \times 0.01 \times (0.9)^{18} \approx 190 \times 0.01 \times 0.1496 \approx 0.2843 \]

### Step 5: Sum the probabilities
Now we can find \( P(X \leq 2) \):
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \leq 2) \approx 0.1216 + 0.2702 + 0.2843 \approx 0.6761 \]

### Step 6: Calculate \( P(X > 2) \)
Now we can find \( P(X > 2) \):
\[ P(X > 2) = 1 - P(X \leq 2) \]
\[ P(X > 2) = 1 - 0.6761 \approx 0.3239 \]

### Final Answer
The probability that more than 2 bulbs are defective is approximately **0.3239**.

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