To solve this problem, we need to calculate the probability that a shipment of 3000 aspirin tablets will be accepted under the given sampling plan. The sampling plan involves testing 42 tablets and accepting the shipment if 1 or fewer of those 42 tablets are defective. The shipment has a 2% defect rate, meaning 2% of the 3000 tablets are defective.

Step 1: Determine the expected number of defective tablets in the sample

We know that 2% of the shipment is defective, so the expected defect rate in the sample of 42 tablets will be 2% of 42.

$\text{Expected defective tablets in sample} = 0.02 \times 42 = 0.84$

Step 2: Model the number of defective tablets

The number of defective tablets in the sample can be modeled by a binomial distribution, where:

• $n = 42$ (the number of tablets sampled),
• $p = 0.02$ (the probability of a defective tablet),
• $k$ is the number of defective tablets found in the sample.

The probability mass function for a binomial distribution is given by:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $X$ is the number of defective tablets in the sample,
• $k$ is the specific number of defective tablets we are interested in (1 or fewer defects),
• $\binom{n}{k}$ is the binomial coefficient.

We are interested in the probability that 1 or fewer tablets are defective, i.e., $P(X \leq 1)$.

Step 3: Calculate the probability $P(X \leq 1)$

We need to compute the probabilities for $X = 0$ and $X = 1$:

• For $X = 0$ (no defective tablets): $P(X = 0) = \binom{42}{0} (0.02)^0 (0.98)^{42} = (0.98)^{42} \approx 0.364$

• For $X = 1$ (one defective tablet): $P(X = 1) = \binom{42}{1} (0.02)^1 (0.98)^{41} = 42 \times 0.02 \times (0.98)^{41} \approx 0.372$

Thus, the total probability that the shipment will be accepted (i.e., 0 or 1 defective tablet) is:

$P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.364 + 0.372 = 0.736$

Step 4: Interpret the result

The probability that the shipment will be accepted is approximately 0.736 or 73.6%.

Conclusion

There is a 73.6% chance that this particular shipment will be accepted under the given acceptance sampling plan.

Given that the defect rate is 2%, it is likely that most shipments with a similar defect rate will be accepted, although some shipments with more than one defective tablet in the sample will be rejected.

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Would you like more details or further clarification? Here are some related questions:

1. How does the binomial distribution change when the defect rate increases to 5%?
2. What would happen to the acceptance probability if the sample size were increased to 50 tablets?
3. What is the impact of reducing the acceptance criterion to 0 defective tablets?
4. How can the probability of acceptance be calculated for a different sampling method, like the hypergeometric distribution?
5. How do you calculate the rejection probability using the complement rule?

Tip: The binomial distribution can be approximated by a normal distribution when $n$ is large and $p$ is not too close to 0 or 1. This is useful for simplifying calculations.