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A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 44 ​tablets, then accept the whole batch if there is only one or none that​ doesn't meet the required specifications. If one shipment of 5000 aspirin tablets actually has a 3% rate of​defects, what is the probability that this whole shipment will be​accepted?

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Final answer:

To calculate the probability of accepting the whole shipment, we need to determine the probability of finding one or no defective tablets in a random sample of 44 tablets. The shipment has a 3% rate of defects, and we can use the binomial probability formula to calculate the probabilities. Finally, we add the probabilities of getting 0 and 1 defective tablets to find the total probability of accepting the shipment.

Step-by-step explanation:

To calculate the probability that the whole shipment will be accepted, we need to determine the probability of finding one or no defective tablets in a random sample of 44 tablets from the shipment.

The shipment has a 3% rate of defects, which means that out of 5000 tablets, 3% or 150 tablets are defective. To find the probability, we can use the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

Where:

  • P(X = k) is the probability of getting exactly k defective tablets
  • C(n, k) is the number of ways to choose k defective tablets from n total tablets
  • p is the probability of a single tablet being defective (3% or 0.03)
  • n is the sample size (44)

In this case, we want to find the probability of getting 0 defective tablets (k = 0) or 1 defective tablet (k = 1). So:

P(X = 0) = C(44, 0) * 0.03^0 * (1-0.03)^(44-0)

P(X = 1) = C(44, 1) * 0.03^1 * (1-0.03)^(44-1)

Finally, we can calculate the total probability of accepting the whole shipment by adding the probabilities of getting 0 and 1 defective tablets:

P(accepting the shipment) = P(X = 0) + P(X = 1)

User Gustavo Ramos
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