Skip to main content

Hypergeometric: solar collectors selection

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.

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.

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).

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

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 2 questions →