---
title: "Binomial: thick coating on 15 chips"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:ic-chips-thick-coating-15"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/ic-chips-thick-coating-15
site: BCT Engineering
---

# Binomial: thick coating on 15 chips

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2076 Asoj, 5 marks
- Video: [Binomial Distribution EXPLAINED in UNDER 15 MINUTES!](https://www.youtube.com/watch?v=e04_wUoscBU) — zedstatistics
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/ic-chips-thick-coating-15
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chips receive a thick enough coating, find the probability that among 15 chips (i) at least 12 will have thick enough coatings; (ii) at most 3 will have thick enough coatings; (iii) exactly 10 will have thick enough coatings.

## Short answer

Using binomial n=15, p=0.7: P(X≥12)=0.2969; P(X≤3)=0.0001; P(X=10)=0.2061.

## Answer

Let X = number of chips with a thick enough coating. n = 15, p = 0.7, q = 0.3, so X ~ B(15, 0.7):

P(X = x) = C(15, x) (0.7)ˣ (0.3)¹⁵⁻ˣ

#### (i) At least 12
```text
P(12) = 455 × 0.7¹² × 0.3³ = 0.17004
P(13) = 105 × 0.7¹³ × 0.3² = 0.09156
P(14) =  15 × 0.7¹⁴ × 0.3  = 0.03052
P(15) =   1 × 0.7¹⁵        = 0.00475
```
P(X ≥ 12) = 0.17004 + 0.09156 + 0.03052 + 0.00475 = **0.2969**

#### (ii) At most 3
```text
P(0) = 0.3¹⁵                = 0.0000000143
P(1) = 15 × 0.7 × 0.3¹⁴     = 0.0000005022
P(2) = 105 × 0.7² × 0.3¹³   = 0.0000082028
P(3) = 455 × 0.7³ × 0.3¹²   = 0.0000829393
```
P(X ≤ 3) = 0.0000917 ≈ **0.0001** (practically zero)

#### (iii) Exactly 10
P(X = 10) = C(15, 10) (0.7)¹⁰ (0.3)⁵ = 3003 × 0.0282475 × 0.00243 = **0.2061**

**Answer: (i) 0.2969, (ii) 0.0001, (iii) 0.2061**

## Check yourself

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

1. The number of well-coated chips among 15 follows:
   A. B(15, 0.3)
   B. B(15, 0.7)
   C. Poisson with lambda 0.7
   D. Negative binomial with k = 15
   Answer: B. Fixed n = 15, independent chips, success p = 0.7. (Key point: Model)

2. P(X >= 12) is computed as:
   A. P(12) + P(13) + P(14) + P(15) = 0.2969
   B. P(0) + ... + P(12)
   C. P(12) only = 0.1700
   D. 1 - P(12) = 0.83
   Answer: A. 'At least 12' means 12, 13, 14 or 15. (Key point: At least 12)

3. P(X <= 3) is practically zero because:
   A. p is very small
   B. n is too large
   C. With p = 0.7, X is almost always much larger than 3
   D. The trials are dependent
   Answer: C. The mean is 10.5, so 3 or fewer is extremely unlikely. (Key point: At most 3)

4. The formula for P(X = 10) is:
   A. C(15,10) (0.3)^10 (0.7)^5
   B. C(10,15) (0.7)^10 (0.3)^5
   C. (0.7)^10 (0.3)^5
   D. C(15,10) (0.7)^10 (0.3)^5
   Answer: D. P = C(n,x) p^x q^(n-x) with x = 10 successes and 5 failures. (Key point: Exactly 10)

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