1. A batch of 10 rocker cover gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement, from the batch of 10, find the probability that a sample contains 2 defective gaskets.
1. A batch of 10 rocker cover gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement, from the batch of 10, find the probability that a sample contains 2 defective gaskets.
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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1. Please solve this Hypergeometric Problem. Please make sure to put all the givens and show the work organized. Thank you.
![### Probability Problem on Defective Gaskets
**Problem Statement:**
A batch of 10 rocker cover gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement from the batch of 10, find the probability that a sample contains 2 defective gaskets.
**Solution Approach:**
To solve this problem, we can use the hypergeometric distribution formula. The hypergeometric distribution describes the probability of k successes in n draws from a finite population of size N that contains exactly K successes, without replacement.
In this context:
- \( N \) is the total number of gaskets, which is 10.
- \( K \) is the number of defective gaskets, which is 4.
- \( n \) is the sample size, which is 3.
- \( k \) is the number of defective gaskets we want in the sample, which is 2.
The probability formula for the hypergeometric distribution is:
\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]
Where \( \binom{n}{r} \) is a binomial coefficient, also represented as "n choose r," and calculated as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
**Calculation of Probabilities:**
1. Calculate \( \binom{4}{2} \):
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
2. Calculate \( \binom{6}{1} \) (since there are 6 non-defective gaskets and we choose 1 from them):
\[ \binom{6}{1} = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6 \]
3. Calculate \( \binom{10}{3} \):
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
Now, substitute these values into the hypergeometric](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58df8008-4ae5-464c-931b-1ae680edb1c4%2Fcf813ebf-1399-4964-8395-bfa4243b9934%2F1rwtuk3_processed.png&w=3840&q=75)
Transcribed Image Text:### Probability Problem on Defective Gaskets
**Problem Statement:**
A batch of 10 rocker cover gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement from the batch of 10, find the probability that a sample contains 2 defective gaskets.
**Solution Approach:**
To solve this problem, we can use the hypergeometric distribution formula. The hypergeometric distribution describes the probability of k successes in n draws from a finite population of size N that contains exactly K successes, without replacement.
In this context:
- \( N \) is the total number of gaskets, which is 10.
- \( K \) is the number of defective gaskets, which is 4.
- \( n \) is the sample size, which is 3.
- \( k \) is the number of defective gaskets we want in the sample, which is 2.
The probability formula for the hypergeometric distribution is:
\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]
Where \( \binom{n}{r} \) is a binomial coefficient, also represented as "n choose r," and calculated as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
**Calculation of Probabilities:**
1. Calculate \( \binom{4}{2} \):
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
2. Calculate \( \binom{6}{1} \) (since there are 6 non-defective gaskets and we choose 1 from them):
\[ \binom{6}{1} = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6 \]
3. Calculate \( \binom{10}{3} \):
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
Now, substitute these values into the hypergeometric
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