Is My Calculation of Tire Fault Probability Correct or Is My Teacher Right?

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Discussion Overview

The discussion revolves around the calculation of the probability of having faulty tires in a set of four brand X tires, where 2% are known to be faulty. Participants explore the correct application of probability theory, particularly the binomial distribution, in this context.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the probability of having 3 faulty tires as (.02^3 * .98) * 4, arguing that combinations of tire positions must be considered.
  • Another participant supports this calculation, suggesting that the teacher's answer may not account for the different configurations of faulty tires.
  • Some participants propose that the teacher's interpretation could imply a specific configuration of faulty tires, which should have been clarified in the problem statement.
  • A later reply mentions the assumption of independence in tire failures and introduces the binomial distribution as a relevant framework for this type of problem.
  • Concerns are raised about potential real-world correlations in tire failures, suggesting that the independence assumption may not hold in practical scenarios.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to the problem, with no consensus reached regarding the teacher's interpretation versus the student's calculation. The discussion remains unresolved regarding the intended meaning of the problem statement.

Contextual Notes

Participants note that the problem may lack clarity regarding whether specific configurations of faulty tires were intended, and there is uncertainty about the independence of tire failures in real-world contexts.

Firefight
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I'll explain the problem in a nutshell.

2% of brand X tires are faulty, 98% aren't.

If you take a car with 4 brand X tires, what is the probability that 3 of them are faulty?

My work: (.02^3 * .98) * 4 = 0.003136%

However, my teacher thought it was just: (.02^3 * .98) = 0.000784%

I believe my teacher left out the combinations that the 3 tires could be in in 4 spaces. ( 4 combinations ) XXXO, XXOX, XOXX, OXXX.

Leaving this out is correct only if it specified which space the 3 faulty ones had to be in, no?

I mentioned this in class and got shot down by the teacher leaving me somewhat frustrated. Can someone else just confirm this? It would mean a lot.
 
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your solution to the way you formulated the problem is correct.
 
I think you are right as well but you should probably ask your teacher if they meant for a specific configuration or whether they allow any of the four.

Even if it is a specific configuration, the teacher still should have mentioned that when writing the question, if the question is as you have posted it to be.
 
Is your teacher Jackson Guo?
 
Could there have been a miscommunication or misunderstanding of the problem? If not, then your answer (and Dickfore's) is correct. The teacher's answer is the probability that 3 out of 3 randomly-selected tires are faulty.

I'm assuming the failure probability for each tire is independent of the failure probability for any other tire. The general formula for problems of this type is given by the binomial distribution. That's almost certainly what would be intended by a textbook problem.

Real tire failures might be correlated - for example, suppose tires with serial numbers 1000-2000 are a "bad batch" which were poorly made and are especially likely to fail. If you bought all 4 tires at the same time, they might all in the bad batch. Non-independence matters a lot in real-life engineering - but in this context, it's probably an irrelevant technicality.
 

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