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A research team of 9 people consists of 4 people from the Boston office, 3 people from the Denver office, and 2 people from the Austin office. If a subcommittee of 4 people is selected at random from the 9 people, what is the probability that the subcommittee includes people from exactly two of the three offices?
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Correct answer
A
The sample space is choosing 4 from 9, which is C(9,4) equals 126.
A subcommittee uses exactly two offices when all 4 members come from the union of two offices but not all from a single one. Handle each office-pair, subtracting the all-one-office groups inside it.
Boston and Denver (7 people): C(7,4) equals 35, minus all-Boston C(4,4) equals 1, minus all-Denver C(3,4) equals 0, giving 34. Boston and Austin (6 people): C(6,4) equals 15, minus all-Boston 1, minus all-Austin C(2,4) equals 0, giving 14. Denver and Austin (5 people): C(5,4) equals 5, minus all-Denver C(3,4) equals 0, minus all-Austin C(2,4) equals 0, giving 5.
So exactly two offices equals 34 + 14 + 5 equals 53, and the probability is 53/126. The answer is A.
You can confirm by completing the partition. Exactly one office: only all-Boston works, C(4,4) equals 1 (Denver and Austin are too small to fill 4 seats). Exactly three offices, a 2-1-1 split: C(4,2)×C(3,1)×C(2,1) equals 36, plus C(4,1)×C(3,2)×C(2,1) equals 24, plus C(4,1)×C(3,1)×C(2,2) equals 12, total 72. The buckets sum to 1 + 53 + 72 equals 126, the full count.
The distractors track specific slips. 54/126 (B) takes 126 - 72 by complement but forgets to also remove the single all-Boston group, landing one too high. 55/126 (C) adds the three pair-union counts (35 + 15 + 5) without the inclusion-exclusion subtraction. 72/126 (D) answers the all-three-offices count instead. 125/126 (E) misreads exactly two as at least two and adds the exactly-three count. Note that the complement route is no faster here, since it still needs the separate all-three-offices count.
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