A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 45 ​tablets, then accept the whole batch if there is only one or none that​ doesn't meet the required specifications. If one shipment of 7000 aspirin tablets actually has a 4​% rate of​ defects, what is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?
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Part 1
The probability that this whole shipment will be accepted is 
  
enter your response here.

We can approach this problem using the binomial distribution, which is ideal when we are dealing with a fixed number of trials (45 in this case), two possible outcomes for each trial (defective or not defective), and a constant probability of success (4% defect rate).

Let's define the variables:

- \( n = 45 \): number of tablets selected and tested
- \( p = 0.04 \): probability that a tablet is defective
- \( q = 1 - p = 0.96 \): probability that a tablet is not defective
- \( X \): the number of defective tablets in the sample of 45

The sampling plan states that the shipment is accepted if there is one or none defective tablet, i.e., \( X \leq 1 \).

We need to calculate \( P(X \leq 1) \), the probability of getting at most 1 defective tablet in the sample.

### Step 1: Binomial probability formula
The probability of getting exactly \( k \) defective tablets out of \( n \) is given by the binomial probability formula:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}
\]

Where \( \binom{n}{k} \) is the binomial coefficient \( \frac{n!}{k!(n-k)!} \).

### Step 2: Calculate \( P(X = 0) \) and \( P(X = 1) \)

- For \( k = 0 \) (no defective tablets):
  \[
  P(X = 0) = \binom{45}{0} \cdot (0.04)^0 \cdot (0.96)^{45} = 1 \cdot 1 \cdot (0.96)^{45} \approx 0.124
  \]

- For \( k = 1 \) (exactly one defective tablet):
  \[
  P(X = 1) = \binom{45}{1} \cdot (0.04)^1 \cdot (0.96)^{44} = 45 \cdot 0.04 \cdot (0.96)^{44} \approx 0.233
  \]

### Step 3: Find \( P(X \leq 1) \)

Now we sum the probabilities of getting 0 or 1 defective tablet:

\[
P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.124 + 0.233 = 0.357
\]

### Conclusion

The probability that the shipment will be accepted is approximately **0.357**, or 35.7%.

### Interpretation

Since the probability of acceptance is only about 35.7%, **many such shipments will likely be rejected**.

Would you like further details on any part of the solution? Here are some related questions:

1. What is the binomial distribution, and why is it used here?
2. How does the probability change if the sample size or defect rate changes?
3. What other statistical distributions could be used for similar problems?
4. What is the role of the binomial coefficient in these calculations?
5. How could we calculate the probability of getting more than one defective tablet?

**Tip**: Always check if conditions match for using the binomial distribution, such as independent trials and a constant probability of success.

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 45 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 7000 aspirin tablets actually has a 4% rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

Calculate the probability of accepting a shipment of aspirin tablets with a 4% defect rate using a binomial distribution approach. This problem uses 45 randomly selected tablets, showing key steps and formulas like P(X ≤ 1) and the binomial probability formula. Learn how to apply the binomial distribution to quality control scenarios. Suitable for Grades 11-12 or college-level probability courses.

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