Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
A company has 4 production sections, viz. S1, S2, S3 and S4 [#permalink]
04 Nov 2004, 17:41
A company has 4 production sections, viz. S1, S2, S3 and S4 which contribute 30%, 20%, 28% and 22% respectively to the total output. It was observed that these sections respectively produced 1%, 2%, 3% and 4% defective units. If a unit is selected at random and found to be defective, what is the probability that the unit so selected came from section S1.
Let us take 100 units , so S1 produces 30 S2 produces 20 S3 produces 28 S4 produces 22
No of defective from S1 = 1% = .3 No of defective from S2 = 2% = .2 No of defective from S3 = 3% = .28 No of defective from S4 = 4% = .22
Total defective = 1
so, prob of defective, piece = 1/100
P(A+B) = P(A/B)*P(B)
30/100 * .3/30 = P(A/B) * 1/100
or P(A/B) = .3
I may be misunderstanding this question but portion in red should be:
No of defective from S2 = 2% = .4
No of defective from S3 = 3% = .84
No of defective from S4 = 4% = .88
Should it not?
Total would then be .3+.4+.84+.88 = 2.42