What Is the Defective Rivet Probability for Aircraft Seams?

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SUMMARY

The discussion focuses on calculating the defective rivet probability for aircraft seams, specifically addressing two scenarios. In the first scenario, with 20% of seams requiring rework, the probability of a defective rivet is determined to be 0.04, or 4%. In the second scenario, to ensure only 10% of seams need reworking, the probability of a defective rivet must be reduced to 0.02, or 2%. The calculations utilize principles of binomial distribution to derive these probabilities.

PREREQUISITES
  • Understanding of binomial distribution
  • Basic probability theory
  • Knowledge of independent events
  • Familiarity with aircraft engineering standards
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  • Study the binomial distribution in depth
  • Explore probability calculations for independent events
  • Research quality control measures in aircraft manufacturing
  • Learn about statistical process control (SPC) techniques
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Aerospace engineers, quality assurance professionals, and anyone involved in aircraft manufacturing and maintenance will benefit from this discussion.

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An aircraft requires [tex]25[/tex] rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with same probability.

(a) If [tex]20 \%[/tex] of all seams need reworking, what is the probability that a rivet is defective?

So this is equaled to [tex]\frac{1}{25}[/tex]?


(b) How small should the probability of a defective rivet be to ensure that only [tex]10 \%[/tex] of all seams need reworking?
 
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Have you learned about binomial distribution yet?
 
figured it out, thanks.
 

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