- #1
Firefight
- 6
- 0
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.
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.