2. In the manufacture of car tires, a particular production process is know to yield 10 tires with defective walls in every batch of 100 tires produced. From a production batch of 100 tires, a sample of 4 is selected for testing to destruction. Find the probability that the sample antning 1 defective tire.
2. In the manufacture of car tires, a particular production process is know to yield 10 tires with defective walls in every batch of 100 tires produced. From a production batch of 100 tires, a sample of 4 is selected for testing to destruction. Find the probability that the sample antning 1 defective tire.
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 11ECP: A manufacturer has determined that a machine averages one faulty unit for every 500 it produces....
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2. Please solve this Hypergeometric Problem. Please make sure to put all the givens and show the work organized. Thank you.
![### Probability of Finding Defective Tires in a Sample
#### Problem Statement:
In the manufacture of car tires, a particular production process is known to yield 10 tires with defective walls in every batch of 100 tires produced. From a production batch of 100 tires, a sample of 4 is selected for testing to destruction. Find the probability that the sample contains 1 defective tire.
#### Explanation:
- **Total tires in the batch:** 100
- **Number of defective tires:** 10
- **Sample size (number of tires selected for testing):** 4
To determine the probability that the sample of 4 tires contains exactly 1 defective tire, we can use the hypergeometric distribution formula.
The hypergeometric distribution is defined as follows:
\[ P(X = k) = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}} \]
Where:
- \( N \) is the population size (100 tires),
- \( D \) is the number of defective items in the population (10 defective tires),
- \( n \) is the sample size (4 tires), and
- \( k \) is the number of defective items in the sample (1 defective tire).
By substituting the given values into the formula, you can compute the exact probability.
### Calculation:
\[ P(X = 1) = \frac{\binom{10}{1} \binom{90}{3}}{\binom{100}{4}} \]
Using combinations:
\[ P(X = 1) = \frac{ \left( \frac{10!}{1!(10-1)!} \right) \left( \frac{90!}{3!(90-3)!} \right) }{ \left( \frac{100!}{4!(100-4)!} \right) } \]
\[ P(X = 1) = \frac{ 10 \cdot \left( \frac{90 \times 89 \times 88}{3 \times 2 \times 1} \right) }{ \frac{100 \times 99 \times 98 \times 97}{4 \times 3 \times 2 \times 1} } \]
After simplifying the above combinations and calculations, you would find the precise probability.
### Conclusion:
The hypergeometric distribution provides a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58df8008-4ae5-464c-931b-1ae680edb1c4%2F6d32f2c0-18f6-435b-8727-c381448fa5ca%2F7c6tkx_processed.png&w=3840&q=75)
Transcribed Image Text:### Probability of Finding Defective Tires in a Sample
#### Problem Statement:
In the manufacture of car tires, a particular production process is known to yield 10 tires with defective walls in every batch of 100 tires produced. From a production batch of 100 tires, a sample of 4 is selected for testing to destruction. Find the probability that the sample contains 1 defective tire.
#### Explanation:
- **Total tires in the batch:** 100
- **Number of defective tires:** 10
- **Sample size (number of tires selected for testing):** 4
To determine the probability that the sample of 4 tires contains exactly 1 defective tire, we can use the hypergeometric distribution formula.
The hypergeometric distribution is defined as follows:
\[ P(X = k) = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}} \]
Where:
- \( N \) is the population size (100 tires),
- \( D \) is the number of defective items in the population (10 defective tires),
- \( n \) is the sample size (4 tires), and
- \( k \) is the number of defective items in the sample (1 defective tire).
By substituting the given values into the formula, you can compute the exact probability.
### Calculation:
\[ P(X = 1) = \frac{\binom{10}{1} \binom{90}{3}}{\binom{100}{4}} \]
Using combinations:
\[ P(X = 1) = \frac{ \left( \frac{10!}{1!(10-1)!} \right) \left( \frac{90!}{3!(90-3)!} \right) }{ \left( \frac{100!}{4!(100-4)!} \right) } \]
\[ P(X = 1) = \frac{ 10 \cdot \left( \frac{90 \times 89 \times 88}{3 \times 2 \times 1} \right) }{ \frac{100 \times 99 \times 98 \times 97}{4 \times 3 \times 2 \times 1} } \]
After simplifying the above combinations and calculations, you would find the precise probability.
### Conclusion:
The hypergeometric distribution provides a
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