# Production line problem

• Sep 4th 2006, 08:00 AM
andy_atw
Production line problem
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%

thanks

Andy
• Sep 4th 2006, 10:25 AM
Quick
although I haven't checked your arithmetic...
• Sep 4th 2006, 11:31 AM
CaptainBlack
Quote:

Originally Posted by andy_atw
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%

Check your arithmetic, and look up the difference between a decimal
fraction and a percentage.

RonL
• Sep 4th 2006, 11:37 AM
CaptainBlack
Quote:

Originally Posted by andy_atw
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%

This part of the question is asking for P(F|E) not P(E|F), and

P(F|E)=P(E and F)/P(E)=0.3333x0.05/0.03 ~=0.555..