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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