Your works correct
although I haven't checked your arithmetic...
I need to know if I am on the right track with this.
A factory has 3 shifts. the shifts produce 1%, 2% and 5% defective items respectively. If all shifts have the same productivity, (1) what percentage of the items produced in a day are defective? (2) what is the probability that the defective items were produced on the third shift?
This is what I came up with:
let n = total number of items produced daily.
each shift produces n/3 items.
we are looking for total (defective/total produced).
=((n/3)(.01+.02+.05))/(n) which equals .03%
then...let E denote the probability that a defective item was produced.
let F denote the probability it was produced on the 3rd shift.
P(E|F)=P(E n F)/P(F) = .03/.05 = .6 or 60%