Questions: A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 43 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 3000 aspirin tablets actually has a 6% 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? The probability that this whole shipment will be accepted is (Round to four decimal places as needed.)

Solution Steps

A pharmaceutical company tests a random sample of 43 aspirin tablets from a shipment of 3000 tablets, which has a defect rate of \( p = 0.06 \). The acceptance criterion is that the shipment will be accepted if there are at most 1 defective tablet in the sample.

To find the probability of exactly 0 defects, we use the binomial probability formula:

\[ P(X = 0) = \binom{n}{0} \cdot p^0 \cdot q^{n-0} = 1 \cdot (0.06)^0 \cdot (0.94)^{43} = 0.0699 \]

Thus, the probability of exactly 0 defects is:

\[ P(X = 0) = 0.0699 \]

Next, we calculate the probability of exactly 1 defect:

\[ P(X = 1) = \binom{n}{1} \cdot p^1 \cdot q^{n-1} = 43 \cdot (0.06)^1 \cdot (0.94)^{42} = 0.1919 \]

Thus, the probability of exactly 1 defect is:

\[ P(X = 1) = 0.1919 \]

The total probability of accepting the shipment is the sum of the probabilities of having 0 or 1 defect:

\[ P(\text{Accept}) = P(X = 0) + P(X = 1) = 0.0699 + 0.1919 = 0.2618 \]

Final Answer