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

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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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