A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 13 ​tablets, then accept the whole batch if there is only one or none that​ doesn't meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 6​% rate of​ defects, what is the probability that this whole shipment will be​ accepted?

Answer:0.818Step-by-step explanation:Since the shipment has a ton of aspirin tablets, we can assume that we pick 13 of them with reposition, because the probability shoudn't change dramatically from the probability of picking without reposition if we do so.We call D the amount of defective tablets. If we assume that we pick the tablets with reposition, then we obtain that D is a random variable of Binomial distribution with parameters 13 and 0.6 (the probability of picking a defective tablet).We want D to be at most one. To calculate the probability of that event we add up the probability of D being equal to 0 and the probability of D being equal to one. Since D is binomial, we have[tex]P(D = 0) = (1-0.06)^{13} = 0.447[/tex][tex]P(D = 1) = {13 \choose 1} * 0.06^1 * (1-0.06)^{12} = 13*0.06*0.94^{12} = 0.371[/tex]We conclude that[tex] P(D \leq 1) = P(D = 0) + P(D = 1) = 0.447+0.371 = 0.818 [/tex]Hence, the shipment will be accepted with probability 0.818I hope this helps you!