Suppose a lot of 25 machine parts is delivered and that a part is considered acceptable only if it passes tolerance test. Suppose further we sample 10 parts and find that none are defective. What is the probability of this event if there are 6 defective parts, that is, parts that do not pass a tolerance test, in a lot of 25?
Suppose a lot of 25 machine parts is delivered and that a part is considered acceptable only if it passes tolerance test. Suppose further we sample 10 parts and find that none are defective. What is the probability of this event if there are 6 defective parts, that is, parts that do not pass a tolerance test, in a lot of 25?
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![**Probability of Sampling Non-Defective Parts**
**Problem Statement:**
Suppose a lot of 25 machine parts is delivered and that a part is considered acceptable only if it passes a tolerance test. Suppose further we sample 10 parts and find that none are defective. What is the probability of this event if there are 6 defective parts, that is, parts that do not pass a tolerance test, in a lot of 25?
**Solution Explanation:**
To determine the probability of sampling 10 parts and finding that none are defective given that there are 6 defective parts in a lot of 25, we need to use the hypergeometric distribution.
The hypergeometric distribution describes the probability of k successes (in this case, non-defective parts) in n draws (the sampled parts), from a finite population of size N that contains exactly K successful objects.
Given:
- Total number of parts, N = 25
- Number of defective parts, K = 6
- Number of non-defective parts, N - K = 25 - 6 = 19
- Sample size, n = 10
- Number of defective parts in the sample, k = 0 (since we found none defective)
The probability is calculated using the hypergeometric formula:
\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]
Where:
- \(\binom{N}{n}\) is the binomial coefficient.
Substitute the values:
\[ P(X = 0) = \frac{\binom{6}{0} \binom{19}{10}}{\binom{25}{10}} \]
This can be computed using binomial coefficients:
\[ \frac{1 \times \binom{19}{10}}{\binom{25}{10}} \]
**Calculations**:
\[ \binom{19}{10} = \frac{19!}{10!(19-10)!} = \frac{19!}{10! \cdot 9!} \]
\[ \binom{25}{10} = \frac{25!}{10!(25-10)!} = \frac{25!}{10! \cdot 15!} \]
**Final Probability**:
\[ P(X = 0) = \frac{\frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45fdd75f-924d-4ac4-9d15-03fc1f26bc02%2Ffb87e8db-c6a9-45c2-b824-323fa1cbf269%2F5go86gc_processed.png&w=3840&q=75)
Transcribed Image Text:**Probability of Sampling Non-Defective Parts**
**Problem Statement:**
Suppose a lot of 25 machine parts is delivered and that a part is considered acceptable only if it passes a tolerance test. Suppose further we sample 10 parts and find that none are defective. What is the probability of this event if there are 6 defective parts, that is, parts that do not pass a tolerance test, in a lot of 25?
**Solution Explanation:**
To determine the probability of sampling 10 parts and finding that none are defective given that there are 6 defective parts in a lot of 25, we need to use the hypergeometric distribution.
The hypergeometric distribution describes the probability of k successes (in this case, non-defective parts) in n draws (the sampled parts), from a finite population of size N that contains exactly K successful objects.
Given:
- Total number of parts, N = 25
- Number of defective parts, K = 6
- Number of non-defective parts, N - K = 25 - 6 = 19
- Sample size, n = 10
- Number of defective parts in the sample, k = 0 (since we found none defective)
The probability is calculated using the hypergeometric formula:
\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]
Where:
- \(\binom{N}{n}\) is the binomial coefficient.
Substitute the values:
\[ P(X = 0) = \frac{\binom{6}{0} \binom{19}{10}}{\binom{25}{10}} \]
This can be computed using binomial coefficients:
\[ \frac{1 \times \binom{19}{10}}{\binom{25}{10}} \]
**Calculations**:
\[ \binom{19}{10} = \frac{19!}{10!(19-10)!} = \frac{19!}{10! \cdot 9!} \]
\[ \binom{25}{10} = \frac{25!}{10!(25-10)!} = \frac{25!}{10! \cdot 15!} \]
**Final Probability**:
\[ P(X = 0) = \frac{\frac{
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