Find the probability.

A company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 30 components and accept the whole batch if there are fewer than 3 defectives. If a particular shipment of thousands of components actually has a 5% rate of defects, what is the probability that this whole shipment will be accepted?

To solve this problem, we need to determine the probability of accepting the entire shipment, given the acceptance sampling plan. The probability of accepting the shipment corresponds to finding the probability that there are fewer than 3 defective components in a sample of 30, given that the defect rate is 5%.

### Step-by-Step Solution

1. **Define the parameters**:
   - Let \( X \) represent the number of defective components in the sample of 30.
   - The shipment has a 5% defect rate, so the probability of selecting a defective component in each trial is \( p = 0.05 \).
   - The sample size is \( n = 30 \).
   - The shipment will be accepted if there are fewer than 3 defectives, i.e., \( X < 3 \).

2. **Identify the probability distribution**:
   - Since we are dealing with a fixed number of trials (30), with two possible outcomes (defective or not defective) and a constant probability of defect (5%), \( X \) follows a binomial distribution: 
     \[
     X \sim \text{Binomial}(n = 30, p = 0.05)
     \]

3. **Calculate the probability of acceptance**:
   - We need to find \( P(X < 3) \), which is the cumulative probability of finding 0, 1, or 2 defective components.

   Using the binomial probability formula:
   \[
   P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
   \]
   where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\binom{n}{k} = \frac{n!}{k!(n - k)!}\).

   **Probability Calculations**:
   1. \( P(X = 0) = \binom{30}{0} (0.05)^0 (0.95)^{30} \)
   2. \( P(X = 1) = \binom{30}{1} (0.05)^1 (0.95)^{29} \)
   3. \( P(X = 2) = \binom{30}{2} (0.05)^2 (0.95)^{28} \)

   Calculate each term and sum them to find \( P(X < 3) \).

#### Step-by-Step Calculation

Let's go through these calculations:

1. **\( P(X = 0) \):**
   \[
   P(X = 0) = (0.95)^{30} \approx 0.2146
   \]

2. **\( P(X = 1) \):**
   \[
   P(X = 1) = 30 \times (0.05) \times (0.95)^{29} \approx 0.3396
   \]

3. **\( P(X = 2) \):**
   \[
   P(X = 2) = \frac{30 \times 29}{2} \times (0.05)^2 \times (0.95)^{28} \approx 0.2787
   \]

Adding these probabilities together:

\[
P(X < 3) \approx 0.2146 + 0.3396 + 0.2787 = 0.8329
\]

Therefore, the probability that the shipment will be accepted is approximately **0.833** or **83.3%**.

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Would you like further details on any of these steps, or do you have any other questions?

### Related Questions:
1. What is the probability of finding exactly 3 defectives in the sample?
2. How would the acceptance probability change if the defect rate was 10% instead of 5%?
3. What is the probability of finding at least one defective in the sample?
4. How does the sample size affect the probability of acceptance given a 5% defect rate?
5. What would be the acceptance probability if the threshold for defectives was raised to fewer than 5?

**Tip:** The binomial distribution is often approximated by the Poisson distribution when the number of trials is large, and the probability of success is small.

A company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 30 components and accept the whole batch if there are fewer than 3 defectives. If a particular shipment of thousands of components actually has a 5% rate of defects, what is the probability that this whole shipment will be accepted?

Find the probability that a shipment with a 5% defect rate will be accepted using an acceptance sampling plan. This detailed solution uses the binomial distribution to calculate the cumulative probability of fewer than 3 defectives in a sample of 30 components. Ideal for advanced high school and undergraduate students.

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