A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 54 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 6000 aspirin tablets actually has a \( 3 \% \) 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.)

To find the probability that the entire shipment will be accepted, we can use the binomial distribution. In this case, we have 54 trials (the number of tablets tested), and we're interested in the probability of finding at most one defective tablet. Given that the defect rate is 3%, the probability of a tablet being defective is \( p = 0.03 \), and the probability of a tablet being non-defective is \( q = 1 - p = 0.97 \). The acceptance criterion is to accept the batch if there are 0 or 1 defective tablets. Therefore, we need to calculate the probabilities for 0 defects and 1 defect: 1. **Probability of 0 defects (k=0)**: \[ P(X=0) = \binom{54}{0} (0.03)^0 (0.97)^{54} = 1 \cdot 1 \cdot (0.97)^{54} \] 2. **Probability of 1 defect (k=1)**: \[ P(X=1) = \binom{54}{1} (0.03)^1 (0.97)^{53} = 54 \cdot (0.03) \cdot (0.97)^{53} \] Now, let's calculate these probabilities using the values provided. Calculating \( (0.97)^{54} \) and \( (0.97)^{53} \): \[ (0.97)^{54} \approx 0.1874 \] \[ (0.97)^{53} \approx 0.1920 \] Now, substituting these back into the probabilities: - For 0 defects: \[ P(X=0) \approx 1 \cdot (0.97)^{54} = 0.1874 \] - For 1 defect: \[ P(X=1) \approx 54 \cdot 0.03 \cdot (0.97)^{53} \approx 54 \cdot 0.03 \cdot 0.1920 \approx 0.3110 \] Now, summing both probabilities to find the total probability of accepting the shipment: \[ P(X \leq 1) = P(X=0) + P(X=1) \approx 0.1874 + 0.3110 \approx 0.4984 \] Thus, the probability that the whole shipment will be accepted is: \[ \boxed{0.4984} \] Since this acceptance probability is around 50%, it suggests that almost half of the shipments may be accepted under these conditions, which is a significant chance of rejection for shipments with a 3% defect rate.