A manufacturer of liquid crystal displays (LCDs) is studying their production lines. The probability of sampling a defective LCD is 0.1. A sample of 5 LCDs is taken. You may assume that an LCD being defective is independent of any of the others being defective. What is the probability that all 5 LCDs are NOT defective?
A manufacturer of liquid crystal displays (LCDs) is studying their production lines. The probability of sampling a defective LCD is 0.1. A sample of 5 LCDs is taken. You may assume that an LCD being defective is independent of any of the others being defective. What is the probability that all 5 LCDs are NOT defective?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Calculating Probability of Non-Defective LCD Samples**
In this problem, we analyze the probability of obtaining non-defective liquid crystal displays (LCDs) from a production line. The probability of sampling a defective LCD is given as 0.1. A sample set of 5 LCDs is taken, and it is assumed that the defectiveness of each LCD is independent of the others.
**Question:**
What is the probability that all 5 LCDs in the sample are NOT defective?
### Options:
1. \(0\)
2. \(0.00001\)
3. \(0.06561\)
4. \(0.32805\)
5. \(0.59049\)
6. \(0.6561\)
7. \(0.91854\)
8. \(1\)
9. \(5\)
### Solution:
To find the required probability, we need to consider the non-defective probability for each LCD in the sample.
Given:
- Probability of defective LCD (\( P(D) \)) = 0.1
- Therefore, probability of non-defective LCD (\( P(N) \)) = 1 - \( P(D) \) = 1 - 0.1 = 0.9
Since the defectiveness of each LCD is independent, we use the multiplication rule for independent events. Thus, the probability that all 5 LCDs are non-defective is:
\[ P(\text{All non-defective}) = P(N) \times P(N) \times P(N) \times P(N) \times P(N) \]
\[ P(\text{All non-defective}) = 0.9^5 \]
Calculating this:
\[ 0.9^5 = 0.59049 \]
Hence, the correct answer to the question is:
**Option 5) 0.59049**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3aad06c9-a061-4a71-9e2d-f01704b3cf27%2F37e985a7-06f8-4edd-afe0-47439d4d0653%2Fjkbua8c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating Probability of Non-Defective LCD Samples**
In this problem, we analyze the probability of obtaining non-defective liquid crystal displays (LCDs) from a production line. The probability of sampling a defective LCD is given as 0.1. A sample set of 5 LCDs is taken, and it is assumed that the defectiveness of each LCD is independent of the others.
**Question:**
What is the probability that all 5 LCDs in the sample are NOT defective?
### Options:
1. \(0\)
2. \(0.00001\)
3. \(0.06561\)
4. \(0.32805\)
5. \(0.59049\)
6. \(0.6561\)
7. \(0.91854\)
8. \(1\)
9. \(5\)
### Solution:
To find the required probability, we need to consider the non-defective probability for each LCD in the sample.
Given:
- Probability of defective LCD (\( P(D) \)) = 0.1
- Therefore, probability of non-defective LCD (\( P(N) \)) = 1 - \( P(D) \) = 1 - 0.1 = 0.9
Since the defectiveness of each LCD is independent, we use the multiplication rule for independent events. Thus, the probability that all 5 LCDs are non-defective is:
\[ P(\text{All non-defective}) = P(N) \times P(N) \times P(N) \times P(N) \times P(N) \]
\[ P(\text{All non-defective}) = 0.9^5 \]
Calculating this:
\[ 0.9^5 = 0.59049 \]
Hence, the correct answer to the question is:
**Option 5) 0.59049**
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