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A system runs three shifts. In a day, 1% of the items produced by the first shift are defective, 2% of the second shift's items are defective, and 5% of the third shift's items are defective. If the shifts all have the same productivity, what percentage of the items produced in a day are defective?
1/100 * 2/100 * 5/100 = 10/100 = 1/10, so 10% percent defective right?
If an item is defective, what is the probability that it was produced by the 3rd shift?
2% + 1% + 5% = 8% total, 5% / 8% = 0.625
Are these answers correct? Thank you for your time
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The second answer is right, but for the first one you need to take an average.
So the answer would be (1% + 2% + 5%)/3 = ~2.6%.
Why did the vector cross the road?
It wanted to be normal.
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Why do you have to get the average? Is it because of the different shifts?? Thanks for replying.
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Imagine that each one shift produces a hundred items. The first shift produced 100 total and 1%=one item from the produced hundred is defective. The second shift: 100 total and 2%=2 items are defective. Same for the third shift. So in total, during the day, all three shifts produced 100*3=300 items. From this 300, there were 1+2+5=8 defective items. 8 items of 300 is 2.(6)%
Or, since all shifts are equal, you can simply take the average of three, but this solution would work only for equal shifts.
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