In a sampling a large number of parts manufactured by a machine, the mean number of defectives in a sample of 20 is 2. Out of 1000 such samples, how many would be expected to contain at least 3 defective parts?

This problem involves the binomial distribution, which models the number of defective parts in each sample. Let's walk through the solution step by step.

### Step 1: Define Variables
- Let \( X \) be the number of defective parts in a sample of 20 parts.
- \( X \) follows a binomial distribution with parameters \( n = 20 \) and \( p \), where \( n \) is the number of trials and \( p \) is the probability of a defective part in each trial.

From the given information:
- The mean number of defectives in each sample of 20 is 2. In a binomial distribution, the mean is given by \( \mu = n \cdot p \), so:
  \[
  \mu = 20 \cdot p = 2
  \]
  Solving for \( p \), we get:
  \[
  p = \frac{2}{20} = 0.1
  \]
  Hence, the probability of a defective part, \( p \), is 0.1.

### Step 2: Set Up the Binomial Distribution
The binomial distribution for the number of defective parts in a sample is:
\[
P(X = k) = \binom{20}{k} \cdot (0.1)^k \cdot (0.9)^{20-k}
\]
Where \( P(X = k) \) is the probability of having exactly \( k \) defective parts in a sample of 20.

### Step 3: Find Probability of at Least 3 Defective Parts
We are asked to find the probability that a sample contains at least 3 defective parts, i.e., \( P(X \geq 3) \). This can be computed as:
\[
P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))
\]
We will now calculate these individual probabilities.

### Step 4: Calculate \( P(X = 0), P(X = 1), P(X = 2) \)
Using the binomial formula:

1. \( P(X = 0) = \binom{20}{0} \cdot (0.1)^0 \cdot (0.9)^{20} = (0.9)^{20} \)
2. \( P(X = 1) = \binom{20}{1} \cdot (0.1)^1 \cdot (0.9)^{19} = 20 \cdot 0.1 \cdot (0.9)^{19} \)
3. \( P(X = 2) = \binom{20}{2} \cdot (0.1)^2 \cdot (0.9)^{18} = \frac{20 \cdot 19}{2} \cdot (0.1)^2 \cdot (0.9)^{18} \)

Let me calculate these values.

The calculated probabilities are as follows:

- \( P(X = 0) = 0.1216 \)
- \( P(X = 1) = 0.2702 \)
- \( P(X = 2) = 0.2852 \)

Thus, the probability that a sample contains at least 3 defective parts is:
\[
P(X \geq 3) = 1 - (0.1216 + 0.2702 + 0.2852) = 0.3231
\]

### Final Answer:
Out of 1000 samples, the expected number of samples containing at least 3 defective parts is:
\[
0.3231 \times 1000 = 323 \text{ samples}
\]

Would you like more details or further clarifications?

Here are some related questions you might find helpful:
1. How is the binomial distribution used in quality control?
2. How can the mean and variance of a binomial distribution be computed?
3. What assumptions are made when using the binomial distribution?
4. What is the significance of "at least" in probability problems?
5. How would this problem change if the sample size increased to 50?

**Tip:** Binomial distribution problems are easier to solve when you use cumulative probability tables or software for large sample sizes!

In a sampling of a large number of parts manufactured by a machine, the mean number of defectives in a sample of 20 is 2. Out of 1000 such samples, how many would be expected to contain at least 3 defective parts?

Find out how to calculate the probability of at least 3 defective parts in a sample of 20 using binomial distribution. This example explores key probability concepts, with a mean number of defectives of 2, and includes a step-by-step solution. Suitable for Grades 10-12.

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