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Sub-group of engineers, doctors, statisticians

Short answer

Total selections from 9 people: C(9,3)=84. (a) One each: 4×3×2=24 ways, probability=24/84=2/7≈0.286. (b) At least one engineer: 1 − C(5,3)/84 = 1 − 10/84 = 74/84 = 37/42≈0.881.

From a group of 4 Engineers, 3 Doctors and 2 Statisticians a sub-group of 3 has to be made, what is the probability that sub-group consists of a) One from each profession b) At least one engineer

There are 4 + 3 + 2 = 9 people. The number of ways to choose any 3 is

C(9,3) = 9!/(3! 6!) = 84

a) One from each profession

Ways = C(4,1) × C(3,1) × C(2,1) = 4 × 3 × 2 = 24
P = 24 / 84 = 2/7 = 0.2857

b) At least one engineer

Use the complement: no engineer means all 3 are chosen from the 5 non-engineers (3 doctors and 2 statisticians).

C(5,3) = 10
P(no engineer) = 10/84
P(at least one engineer) = 1 − 10/84 = 74/84 = 37/42 = 0.881

Answer: (a) 2/7 ≈ 0.286, (b) 37/42 ≈ 0.881.

✓ Answer checked by Claude Sonnet 5.5

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