--- title: "Hypergeometric: solar collectors selection" subject: "Probability and Statistics" chapter: 2 key: "semester_5/probability_and_statistics:solar-collectors-hypergeometric" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/solar-collectors-hypergeometric site: BCT Engineering --- # Hypergeometric: solar collectors selection Probability and Statistics · Chapter 2: Discrete Probability Distribution - Asked in: 2069 Chaitra, 2+2+2 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/solar-collectors-hypergeometric - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2 ## Question Among the 12 solar collectors on display at a trade show, 9 are flat-plate collectors and the others are concentrating collectors. If a person visiting the show randomly selects 6 of the solar collectors to check out, what is the probability that i) None of them will be flat-plate collectors. ii) At least 3 of them will be flat-plate collectors. iii) At most 2 of them will be concentrating collectors. ## Short answer Hypergeometric with N=12, K=9, n=6. Probability of no flat-plate is 0; at least three flat-plate is 1; at most two concentrating equals at least four flat-plate, probability 10/11 or 0.9091. ## Answer N = 12 collectors: 9 flat-plate and 3 concentrating. A person selects n = 6 without replacement. Let X = number of flat-plate collectors; X is **hypergeometric**: P(X = x) = C(9, x) · C(3, 6 − x) / C(12, 6), C(12, 6) = 924. Only 3 concentrating collectors exist, so at least 6 − 3 = 3 of the six chosen must be flat-plate (x = 3, 4, 5, 6). ```text x C(9,x)·C(3,6-x) P(X = x) 3 84 × 1 = 84 0.0909 4 126 × 3 = 378 0.4091 5 126 × 3 = 378 0.4091 6 84 × 1 = 84 0.0909 ``` #### i) None flat-plate This needs 6 concentrating collectors, but only 3 exist, so P(X = 0) = **0**. #### ii) At least 3 flat-plate Every possible selection has x ≥ 3, so P(X ≥ 3) = 84/924 + 378/924 + 378/924 + 84/924 = 924/924 = **1**. #### iii) At most 2 concentrating This means at most 2 concentrating, i.e. at least 4 flat-plate: P(X ≥ 4) = (378 + 378 + 84) / 924 = 840 / 924 = 10/11 = **0.9091** **Answer: (i) 0, (ii) 1, (iii) 0.9091** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The number of flat-plate collectors chosen follows: A. Binomial with n = 6, p = 0.75 B. Poisson C. Hypergeometric with N = 12, K = 9, n = 6 D. Negative binomial Answer: C. Selection is without replacement from a finite set. (Key point: Model) 2. P(no flat-plate) is 0 because: A. Flat-plate collectors are more common B. C(12,6) is too large C. Selection is with replacement D. Only 3 concentrating collectors exist, but 6 would be needed Answer: D. Six non-flat-plate picks are impossible with only 3 of them. (Key point: None flat-plate) 3. 'At most 2 concentrating' translates to: A. At most 2 flat-plate B. At least 4 flat-plate, giving 10/11 C. Exactly 2 flat-plate D. At least 3 flat-plate Answer: B. 6 picks with at most 2 concentrating means 4, 5 or 6 flat-plate: 840/924. 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