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# A company has 4 production sections, viz. S1, S2, S3 and S4

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Manager
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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.
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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

= 30%
GMAT Club Legend
Joined: 15 Dec 2003
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PracticeMore wrote:
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

= 30%

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
.3/2.42=12.4%
_________________

Best Regards,

Paul

Manager
Joined: 18 Sep 2004
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Location: Dallas, TX
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15/121

Let's say there are 10,000 units produced to get rid of the fractions.
# of units
S1 - 3000
S2 - 2000
S3 - 2800
S4 - 2200

Total units = 10,000

# of defects
S1 - 30
S2 - 40
S3 - 84
S4 - 88

Total defects = 242

If one of the units is defective, then there is a 30/242 chance it came from S1, reduce the fraction and you get 15/121.
Director
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I am also getting 15/121 (or 30/242 or 12.4%)

Last edited by venksune on 05 Nov 2004, 06:33, edited 1 time in total.
Manager
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Quote:
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

Sorry guys... that's a blunder... I hate this type of mistake

No of defective from S1 = 1% = .3
No of defective from S2 = 2% = .4
No of defective from S3 = 3% = .84
No of defective from S4 = 4% = .88

And so equation should be

30/100 * .3/30 = P(A/B) * 2.42/100

so P(A/B) = .3/2.42 = 12.4 %

Thanks Paul for correcting me.

I think it will be good if I validate it once the easier way, ie try conditional probability and validate it by the easier way.
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