Ref : Design for Six Sigma Statistics (3.1.4.2)
Suppose "n" items selected randomly from a finite population of "N" items
If "D" = defective items in a population
Then, Non-defective items in the population = N-D
What is the probabilities samples will contain exactly "x" defective items and (n-x) non-defective items ?
Suppose "n" items selected randomly from a finite population of "N" items
If "D" = defective items in a population
Then, Non-defective items in the population = N-D
What is the probabilities samples will contain exactly "x" defective items and (n-x) non-defective items ?
Example
Suppose a box contain 50 parts
Based on historical data, average defective part found per box = 4% (2 defective part per box)
If each assembly required 5 part pick up randomly from the box, what would be the probability of assembly containing 1 or more defective part ?
This is a finite population sampling problem, where the population = 50pcs (including 2 defective parts)
There are 2 possible outcome from this case
1. assembly containing 1 defective part
2. assembly containing 2 defective part
Since these are mutually exclusive events, the probabilities add.
Solution as below



















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