---
title: "Sub-group of engineers, doctors, statisticians"
subject: "Probability and Statistics"
chapter: 1
key: "semester_5/probability_and_statistics:engineers-doctors-statisticians-subgroup"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/engineers-doctors-statisticians-subgroup
site: BCT Engineering
---

# Sub-group of engineers, doctors, statisticians

Probability and Statistics · Chapter 1: Descriptive Statistics and Basic Probability

- Asked in: 2074 Chaitra, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/engineers-doctors-statisticians-subgroup
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1

## Question

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

## 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.

## Answer

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

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

#### a) One from each profession

```text
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).

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How many ways to choose 3 from 9?
   A. C(9,3) = 84
   B. 9 x 8 x 7 = 504
   C. 9^3 = 729
   D. 3
   Answer: A. Order does not matter, so combinations. (Key point: Total ways)

2. What is P(one from each profession)?
   A. 1/84
   B. 4x3x2/84 = 2/7
   C. 4+3+2 = 9/84
   D. 24/9
   Answer: B. 24 favourable ways over 84. (Key point: One from each)

3. How is P(at least one engineer) found?
   A. C(4,1)/84
   B. C(4,3)/84
   C. 1 - C(5,3)/84 = 37/42
   D. 1 - C(4,3)/84
   Answer: C. Complement: all 3 from the 5 non-engineers. (Key point: At least one engineer)

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