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A pharmaceutical company receives large shipments of ibuprofen tablets and uses an acceptance sampling plan. This plan randomly selects and tests 29 tablets, then accepts the whole batch if there is at most one that doesn't meet the required specifications. What is the probability that this whole shipment will be accepted if a particular shipment of thousands of ibuprofen tablets actually has a 15% rate of defects?

User Egis
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Answer:

Explanation:

This problem can be solved using the binomial distribution, which gives the probability of obtaining a certain number of successes in a fixed number of independent trials.

Let p be the probability that a single ibuprofen tablet has a defect, which is given as 15% or 0.15. Then, the probability that a single ibuprofen tablet does not have a defect is 1 - p = 0.85.

The acceptance sampling plan requires that at most one tablet does not meet the required specifications out of 29 tablets. This means that the shipment will be accepted if there are 0 or 1 defective tablets in the sample of 29.

The probability of getting exactly k defective tablets in a sample of n tablets is given by the binomial probability formula:

P(k) = (n choose k) * p^k * (1 - p)^(n - k)

where (n choose k) = n! / (k! * (n - k)!) is the number of ways to choose k defective tablets out of n, and ! denotes the factorial function.

To find the probability that the whole shipment will be accepted, we need to find the probability that there are 0 or 1 defective tablets in the sample of 29:

P(0 or 1 defects) = P(0 defects) + P(1 defect)

= (29 choose 0) * 0.15^0 * 0.85^29 + (29 choose 1) * 0.15^1 * 0.85^28

≈ 0.1098

Therefore, the probability that the whole shipment will be accepted is approximately 0.1098 or 10.98%

User Leetwinski
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