---
title: "Three students solving a problem"
subject: "Probability and Statistics"
chapter: 1
key: "semester_5/probability_and_statistics:three-students-solve-problem"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/three-students-solve-problem
site: BCT Engineering
---

# Three students solving a problem

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

- Asked in: 2075 Asoj, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/three-students-solve-problem
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1

## Question

A problem in statistics is given to three students A, B and C whose chances of solving are 1/2, 3/4 and 1/4 respectively. If all of them try independently, what is the probability that
a) at least one of them will solve it
b) none of them can solve it
c) exactly two of them can solve it

## Short answer

P(A)=1/2, P(B)=3/4, P(C)=1/4. (a) At least one solves: 1 − (1/2×1/4×3/4)=29/32=0.906. (b) None solves: (1/2×1/4×3/4)=3/32=0.094. (c) Exactly two: (AB not C) + (AC not B) + (BC not A) = 9/32+1/32+3/32=13/32≈0.406.

## Answer

Let A, B, C be the events that the student solves the problem.

P(A) = 1/2, P(B) = 3/4, P(C) = 1/4, so P(A′) = 1/2, P(B′) = 1/4, P(C′) = 3/4. The students try independently, so probabilities multiply.

#### a) At least one solves it

```text
P = 1 − P(none solves)
  = 1 − (1/2)(1/4)(3/4)
  = 1 − 3/32
  = 29/32 = 0.906
```

#### b) None can solve it

```text
P = P(A′) P(B′) P(C′) = (1/2)(1/4)(3/4) = 3/32 = 0.094
```

#### c) Exactly two solve it
Three cases: AB solve and C does not, AC solve and B does not, BC solve and A does not.

```text
A, B, not C: (1/2)(3/4)(3/4) = 9/32
A, C, not B: (1/2)(1/4)(1/4) = 1/32
B, C, not A: (1/2)(3/4)(1/4) = 3/32
Total = 13/32 = 0.406
```

**Answer:** (a) 29/32 = 0.906, (b) 3/32 = 0.094, (c) 13/32 = 0.406.

## Check yourself

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

1. Why can probabilities be multiplied?
   A. Events are exclusive
   B. Events are exhaustive
   C. Students try independently
   D. They sum to 1
   Answer: C. Independent events: product rule. (Key point: Independence)

2. What is P(at least one solves)?
   A. 3/32
   B. 1/2 + 3/4 + 1/4
   C. 13/32
   D. 1 - 3/32 = 29/32
   Answer: D. Complement of none solving. (Key point: At least one)

3. What is P(none)?
   A. (1/2)(1/4)(3/4) = 3/32
   B. 9/32
   C. 13/32
   D. 29/32
   Answer: A. Multiply the failure probabilities. (Key point: None)

4. What is P(exactly two)?
   A. 9/32
   B. 9/32 + 1/32 + 3/32 = 13/32
   C. 3/32
   D. 29/32
   Answer: B. Sum three cases of two solvers and one failing. (Key point: Exactly two)

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