Binomial Probability problem.

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Homework Help Overview

The problem involves determining the probability of finding the first defective engine within a specified range of trials, given a defect rate of 10% for engines on an assembly line. The subject area pertains to probability distributions, specifically binomial and geometric distributions.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the application of binomial and geometric distributions to the problem, with some suggesting that the first defective engine's probability can be expressed as a sum of probabilities for trials ranging from the 5th to the 25th. Questions arise regarding the correct formulation and whether to use binomial or geometric distribution.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of geometric distribution, but there is no explicit consensus on the best approach yet.

Contextual Notes

Participants are considering the implications of the defect rate and the range of trials specified in the problem. There is an acknowledgment of potential confusion between the binomial and geometric distributions in this context.

TheHamburgler1
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Homework Statement


10% of engines manufactured on an assembly line are defective. If engines are randomly selected one at a time and tested, what is the probability that the first defective engine will be found between the 5th trial and the 25th trial, inclusive?


Homework Equations





The Attempt at a Solution


I believe this is just a binomial distribution with Bin(n,1) where n varies between 5 and 25.

\sum(nC1)(0.1)(.9)^(n-1)

This is way off because I am getting 6.47...

thanks ahead of time.
 
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Hi TheHamburgler1! :smile:

No, it's just P(first is defective is the 5th) + … + P(first is defective is the 25th) :wink:
 
tiny-tim said:
Hi TheHamburgler1! :smile:

No, it's just P(first is defective is the 5th) + … + P(first is defective is the 25th) :wink:

Would we not express each of those via Bin(n,.1) where x=1? If not, how would we express one of them?

Thanks
 
Actually, this could be a Geometric distribution problem right? In that case we would sum x from 5 to 25 of (.1)(.9)^(x-1). This gives 0.5843102
 
(just got up :zzz: …)
TheHamburgler1 said:
Actually, this could be a Geometric distribution problem right? In that case we would sum x from 5 to 25 of (.1)(.9)^(x-1).

Yup! :biggrin:

(and I assume you've used algebra to calculate that, and not 20 additions? :wink:)
 

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