A pharmaceutical company receives large shipments of ibuprofen tablets and uses an acceptance sampling plan. This plan randomly selects and tests 30 tablets, then accepts the whole batch if there is at most one that doesn't metet 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 $1 \%$ rate of defects?
(Report answer as a decimal value accurate to four decimal places.)
Answer
The probability that the shipment will be accepted is \(\boxed{0.9613}\).
Step 1 ：First, we need to understand the problem. The problem is asking for the probability that a shipment of ibuprofen tablets will be accepted given that the actual defect rate is \(1 \%\). The shipment will be accepted if there is at most one defective tablet in a random sample of 30 tablets.
Step 2 ：We can use the binomial probability formula to solve this problem. The binomial probability formula is given by:
Step 3 ：\[P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\]
Step 4 ：where \(P(X=k)\) is the probability of \(k\) successes in \(n\) trials, \(C(n, k)\) is the number of combinations of \(n\) items taken \(k\) at a time, \(p\) is the probability of success on a single trial, and \((1-p)\) is the probability of failure on a single trial.
Step 5 ：In this case, we want to find the probability that there are 0 or 1 defective tablets in a sample of 30, so we need to calculate \(P(X=0)\) and \(P(X=1)\) and add them together.
Step 6 ：First, let's calculate \(P(X=0)\):
Step 7 ：\[P(X=0) = C(30, 0) * (0.01^0) * ((1-0.01)^(30-0))\]
Step 8 ：\[P(X=0) = 1 * 1 * (0.99^30)\]
Step 9 ：\[P(X=0) = 0.7397\]
Step 10 ：Next, let's calculate \(P(X=1)\):
Step 11 ：\[P(X=1) = C(30, 1) * (0.01^1) * ((1-0.01)^(30-1))\]
Step 12 ：\[P(X=1) = 30 * 0.01 * (0.99^29)\]
Step 13 ：\[P(X=1) = 0.2216\]
Step 14 ：Finally, we add \(P(X=0)\) and \(P(X=1)\) together to get the total probability that the shipment will be accepted:
Step 15 ：\[P(X=0 \text{ or } X=1) = P(X=0) + P(X=1)\]
Step 16 ：\[P(X=0 \text{ or } X=1) = 0.7397 + 0.2216\]
Step 17 ：\[P(X=0 \text{ or } X=1) = 0.9613\]
Step 18 ：\[\boxed{\text{{Final Answer: }}}\]
Step 19 ：The probability that the shipment will be accepted is \(\boxed{0.9613}\).
