Question

a pharmaceutical company receives large shipments of aspirin tablets. the acceptance sampling plan is to randomly select and test 57 tablets, then accept the whole batch if there is only one or none that doesnt meet the required specifications. if one shipment of 3000 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.)

Explanation:

Step1: Identify the binomial parameters

The binomial formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 57\) (number of trials, i.e., number of tablets tested), \(p=0.03\) (probability of a defective tablet), and we want to find \(P(X\leq1)=P(X = 0)+P(X = 1)\). The combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Calculate \(P(X = 0)\)

For \(k = 0\), \(C(57,0)=\frac{57!}{0!(57-0)!}=1\). Then \(P(X = 0)=C(57,0)\times(0.03)^{0}\times(1 - 0.03)^{57-0}=1\times1\times(0.97)^{57}\). Using a calculator, \((0.97)^{57}\approx0.1790\).

Step3: Calculate \(P(X = 1)\)

For \(k = 1\), \(C(57,1)=\frac{57!}{1!(57 - 1)!}=\frac{57!}{1!×56!}=57\). Then \(P(X = 1)=C(57,1)\times(0.03)^{1}\times(0.97)^{56}=57\times0.03\times(0.97)^{56}\). \((0.97)^{56}\approx0.1845\), so \(P(X = 1)=57\times0.03\times0.1845 = 57\times0.005535=0.3155\).

Step4: Calculate \(P(X\leq1)\)

\(P(X\leq1)=P(X = 0)+P(X = 1)=0.1790+0.3155 = 0.4945\).

Answer:

\(0.4945\)