Probability of Defective Bulbs in Sample Selection: A Case Study on Manufacturing Deficiency
Understanding the probability of selecting defective bulbs from a mixed batch is crucial for quality assurance in manufacturing. This article explores the detailed process of calculating the probability of selecting exactly one defective bulb from a sample of three bulbs, given a specific number of defective bulbs within a set of total bulbs.
Introduction to the Problem
Imagine a scenario where 15 bulbs are selected from a production batch, consisting of 5 defective bulbs and 10 non-defective bulbs. We want to determine the probability of selecting exactly one defective bulb in a sample of three bulbs.
Step-by-Step Solution
The solution to this problem involves the use of combinations and the hypergeometric distribution.
Step 1: Define the Parameters
Total bulbs (N) 15 Defective bulbs (D) 5 Non-defective bulbs (N - D) 10 Bulbs chosen (n) 3Given these parameters, we can proceed to calculate the probability of selecting exactly one defective bulb in a sample of three.
Step 2: Calculate the Total Ways to Choose 3 Bulbs from 15
The total number of ways to choose 3 bulbs from 15 is given by the combination formula:
(binom{15}{3}) (frac{15!}{3!(15-3)!}) (frac{15 times 14 times 13}{3 times 2 times 1}) 455
Step 3: Calculate the Ways to Choose 1 Defective and 2 Non-Defective Bulbs
Step 3.1: Choose 1 Defective Bulb from 5
(binom{5}{1}) (frac{5!}{1!(5-1)!}) 5
Step 3.2: Choose 2 Non-Defective Bulbs from 10
(binom{10}{2}) (frac{10!}{2!(10-2)!}) (frac{10 times 9}{2 times 1}) 45
Step 4: Combine the Results
The total number of ways to choose 1 defective and 2 non-defective bulbs is:
(binom{5}{1} times binom{10}{2}) 5 × 45 225
Step 5: Calculate the Probability
The probability (P) that exactly one of the three bulbs is defective is the ratio of the number of favorable outcomes to the total outcomes:
(P_{1text{ defective}}) (frac{text{Number of ways to choose 1 defective and 2 non-defective}}{text{Total ways to choose 3 bulbs}}) (frac{225}{455})
Step 6: Simplify the Fraction
To simplify (frac{225}{455}):
225 15 × 15 455 5 × 91 5 × 7 × 13The greatest common divisor is 5:
(frac{225 div 5}{455 div 5}) (frac{45}{91})
Final Answer: The probability that exactly one of the three bulbs chosen is defective is:
(boxed{frac{45}{91}})
Factories and Manufacturing Distribution
Consider another scenario in a manufacturing environment where two factories, Factory A and Factory B, produce electric bulbs. Factory A produces 8 defective bulbs and Factory B produces 10 defective bulbs. Factory B produces three times as many bulbs as Factory A each week. If a bulb is chosen at random from the weekly production, what is the probability that the bulb is satisfactory?
Analysis and Solution
Let's define the parameters:
Bulbs from Factory A (FA) 8 Bulbs from Factory B (FB) 10 Total bulbs (TB) FA FB 18The ratio of defective to non-defective bulbs in each factory is:
Factory A: 8 defective / 18 total 8/18 Factory B: 10 defective / 18 total 10/18Given these ratios, the probability of selecting a satisfactory (non-defective) bulb from each factory is:
Factory A: 1 - 8/18 10/18 Factory B: 1 - 10/18 8/18Since Factory A and Factory B produce the same number of bulbs each week, the probability of selecting a bulb from Factory A or Factory B is 1/2.
The overall probability of selecting a satisfactory bulb is:
(P_{satisfactory} P_{FA} times P_{satisfactory|FA} P_{FB} times P_{satisfactory|FB})
(frac{1}{2} times frac{10}{18} frac{1}{2} times frac{8}{18})
(frac{1}{2} times left(frac{10 8}{18}right))
(frac{1}{2} times frac{18}{18})
(frac{1}{2})
The probability that the bulb is satisfactory is (frac{1}{2}).
Final Answer: The probability that the bulb is satisfactory is (boxed{frac{1}{2}})
In conclusion, the methods and principles of probability and combinatorics are essential for understanding and solving such problems in quality assurance and manufacturing. Understanding these concepts can help in improving product quality and ensuring customer satisfaction.