---
title: "Hypergeometric: 3 batteries from 24"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:batteries-3-of-24-defect"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/batteries-3-of-24-defect
site: BCT Engineering
---

# Hypergeometric: 3 batteries from 24

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2074 Chaitra, 6 marks; 2070 Asar, 5 marks (asked 2 times)
- Video: [An Introduction to the Hypergeometric Distribution](https://www.youtube.com/watch?v=L2KMttDm3aY) — jbstatistics
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/batteries-3-of-24-defect
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

A quality control engineers inspects a random sample of 3 batteries from each lot of 24 car batteries that is ready to shipment. If such a lot contain six batteries with slight defects, what is the probabilities that the inspector's sample will contain
a) None of the batteries with defect
b) Only one of the batteries with defect
c) At least two of the batteries with defect

## Short answer

Use hypergeometric distribution: (a) 0.4032, (b) 0.4536, (c) 0.1433.

## Answer

Lot size N = 24, defective K = 6, good = 18, sample n = 3 drawn without replacement, so X (number of defectives in the sample) is **hypergeometric**:

P(X = x) = C(6, x) · C(18, 3 − x) / C(24, 3), where C(24, 3) = 2024.

```text
x   C(6,x)·C(18,3-x)      P(X = x)
0   1 × 816   = 816       0.4032
1   6 × 153   = 918       0.4536
2   15 × 18   = 270       0.1334
3   20 × 1    = 20        0.0099
```

#### a) None defective
P(X = 0) = 816 / 2024 = **0.4032**

#### b) Only one defective
P(X = 1) = 918 / 2024 = **0.4536**

#### c) At least two defective
P(X ≥ 2) = P(2) + P(3) = (270 + 20) / 2024 = 290 / 2024 = **0.1433**

(Check: 1 − 0.4032 − 0.4536 = 0.1432, the same up to rounding.)

**Answer: (a) 0.4032, (b) 0.4536, (c) 0.1433**

## Check yourself

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

1. Why is X hypergeometric?
   A. Rare events in time
   B. Sampling without replacement from a finite lot
   C. Waiting for the first defect
   D. Sampling with replacement
   Answer: B. Draws are dependent and the lot is finite. (Key point: Distribution)

2. What is P(X=0)?
   A. 270/2024
   B. 918/2024
   C. 20/2024
   D. 816/2024 = 0.4032
   Answer: D. C(6,0)C(18,3)/C(24,3) = 816/2024. (Key point: P(none))

3. What is P(X=1)?
   A. 0.0099
   B. 0.1334
   C. 0.4536
   D. 0.4032
   Answer: C. 6x153 = 918, so 918/2024 = 0.4536. (Key point: P(one))

4. How is P(X>=2) found?
   A. P(2) only
   B. 1 - P(3)
   C. P(2)+P(3) = 290/2024 = 0.1433
   D. P(0)+P(1)
   Answer: C. Add 270 and 20 over 2024. (Key point: At least two)

## Close variants set in other exams

- [Hypergeometric: 4 batteries from 24](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abatteries-4-of-24-defect) — Probability and Statistics, chapter 2, asked 4×