Sampling faulty bottles under a penalty contract
A craft brewery in Lugo prepares 12 special bottles for a restaurant. Unknown to the brewer, exactly 3 of the 12 bottles have faulty caps and will spoil. The restaurant receives 4 bottles chosen uniformly at random from the 12. The contract says: if none of the 4 is faulty, the brewery receives 120 euros; if exactly 1 is faulty, the brewery receives 60 euros but pays 20 euros for replacing that bottle; if 2 or more are faulty, the restaurant rejects the delivery, the brewery receives nothing and pays a 200 euro penalty. The brewer can instead pay 45 euros for a perfect sorting that removes all 3 faulty bottles before the restaurant chooses 4 from the remaining 9. The brewer wants the probability of rejection, the expected net income without sorting, and a comparison with sorting. The restaurant also asks for the probability of rejection given that at least one faulty bottle was found.
1. Let X be the number of faulty bottles in the restaurant's sample of 4. Find the probability distribution of X.
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Total choices C(12,4)=495. For k faulty, choose k from 3 and 4-k from 9: P(X=k)=C(3,k)C(9,4-k)/495. Values: P(0)=C(3,0)C(9,4)/495=126/495≈0.2545; P(1)=384/495=252/495≈0.5091; P(2)=336/495=108/495≈0.2182; P(3)=1*9/495=9/495≈0.0182. The sum is 495/495=1.
2. Compute the expected net income without sorting and compare with the 45-euro perfect sorting.
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Contract outcomes: 0 faulty gives +120; 1 faulty gives 60-20=+40; 2 or more gives -200. E=120*(126/495)+40*(252/495)-200*(117/495). The numerator is 15120+10080-23400=1800, so E=1800/495≈3.64 euros. With perfect sorting, the 3 faulty bottles are removed, the restaurant chooses 4 of 9 good bottles and pays 120; net is 120-45=75 euros. Sorting is far better: 75 versus about 3.64, a gain of about 71.36 euros. The brewer should sort unless there are hidden costs or the sorting is imperfect.
3. Find P(X ≥ 2 | X ≥ 1). Interpret it for the restaurant's rejection rule.
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P(X ≥ 2 | X ≥ 1)=P(X ≥ 2)/P(X ≥ 1). P(X ≥ 2)=(108+9)/495=117/495. P(X ≥ 1)=(252+108+9)/495=369/495. Ratio=117/369≈0.317. So once at least one faulty bottle is found, there is about a 31.7% chance that the delivery will be rejected under the 2-or-more rule. This conditional probability is useful because a single faulty bottle is not yet a rejection, but it raises the risk substantially.