Q#7: Defective gaskets

Suppose you’re working in a parts manufacturing plant, and you’re running quality analysis on the gasket production line. Gaskets produced by your company will be defective 1% of the time, and are distributed in packs of 20.

Your company has a policy where if 2 or more of the 20 gaskets in a given pack is defective, they will replace the entire package. What proportion of packages does the company need to replace?

-Credit to: erik@interviewqs.com

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ANSWER:

This question requires a revisit to basic statistics and probability, specifically the binomial distribution.

Using the formula, we see that n=20 (the number of gaskets per package), p = 0.01 for probability of broken gasket, and q = .99 for chance of fixed gasket. X is a little more tricky to place. So our goal is to find the probability of having 2 or more defective gaskets (thus we will have to replace the package). Another way to think about this is that we will have to replace if we dont have 0 broken or 1 broken gasket. Thus the final equation will be written as follows:

P(x >= 2) = 1 - P(x=0) - P(x=1)
Using the binomial formula:
P(x >=2) = 0.0168 (1.68%)