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A shirt factory has a six-stage manufacturing process. Each [#permalink]
 24 Aug 2006, 03:53
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A shirt factory has a six-stage manufacturing process. Each of the six stages is manned by experts and novices, whose sewing times differ, as shown above. A shirt being made passes through each of the six stages, and at each stage, the probability that a shirt is assigned to an expert worker is 0.5. Which most closely approximates the probability that a shirt that is manufactured will undergo at most two hours of sewing?
(A) 0.16 (B) 0.18 (C) 0.20 (D) 0.23 (E) 0.25
Last edited by kevincan on 24 Aug 2006, 13:22, edited 1 time in total.
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Could you please explain how you got the answer to this question ?
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Senior Manager
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You will be able to finish the shirt under two hours if you use a skilled worker in stage-2 and stage-3. It doesn't matter which worker does the job at other stages. The required probability is 0.5*0.5 = 0.25.
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I'm sorry, i messed up.
If novices work on the shirt in all stages, it would take 190 minutes and if experts work on the shirt in all stages, it would take 98 minutes. We can see that the stage-3 is very much critical. If we use a novice in stage-3, there is no way we can complete the shirt in less than or equal to 120 minutes. So, we have
XXEXXX where X denotes either novice or expert. Now, besides the third stage, we need the experts in three more stages. So, the idea is that we must have the expert take care of the stage-3, and also take care of 3 or more stages of the remaining 5. The required probability is
(5C3 + 5C4 + 5C5)/2exp(6) = 0.25
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for every person who doesn't try because he is
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Senior Manager
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Kevin, i had miscalculated the number of minutes. I was under the impression that it will boil down to only two stages. The new answer/explanation has already been posted.
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for every person who doesn't try because he is
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Taking a crack at it [#permalink]
29 Aug 2006, 10:53
This is a tough one.... how much time would you allocate to this?? Took me at least 5 mins to even think through!!! And I'm still not at the answer - this is a partial answer looking for help (I suck at probability and trying to work at it)
Here is what I did:
Total cases are of course 2 ^ 6 = 64.
Min time calculation (all experts) = 99 (best case scenario)
Time diff calculation at each stage = 15, 7, 30, 20, 10, 10.
Now starting from the minimum and seeing which ones have to be done by experts or can be left to the others... to reach a max of 120 (2 hrs).
The first stage, second and last two can be done by expert or non expert (as long as the others are all done by experts). Also 2 and 5 or 2 and 6 can be done togehter by non experts. Any other cases and we wont keep to the time limit.
At this point I'm at a loss how to convert these cases into numbers! But will keep plugging away and come back with a solution if I can.
MG
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Therefore cases are [#permalink]
29 Aug 2006, 10:57
Options are:
1. EEEEEE
2. NEEEEE
3. ENEEEE
4. EEEENE
5. EEEEEN
6. ENEENE
7. ENEEEN
Thats it, pretty much.
7/64 = 0.10
Too low for the answers - back to the drawing board, take 3.
MG
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oooh calculation mistake [#permalink]
29 Aug 2006, 11:08
The summing up was wrong... the expert total actually comes to 97 minutes, leaving us a leeway of 23 minutes and not 21 like I had assumed (infact I somehow managed to assume 19 mins... but this time I have it!!!)
So we can afford anything that will add under 23 mins to the process.
This will add the following cases:
8. NNEEEE (add 22 mins)
9. EEENEE (add 20 mins)
10.EEEENN (add 20 mins)
Thats it!
so its 10/64 = 0.16.
Is that it??
MG
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Full version summary [#permalink]
29 Aug 2006, 11:15
I just wanted to clarify since previous posts were a bit of a working mess :D
Total cases = 64.
Number of cases under 120 mins is what we need to find.
Cases where all Experts work on it = 15+8+25+5+25+20 = 97 mins. This is the minimum time it will take in the case EEEEEE.
We now have a leeway of 23 minutes for novices to work on the system (120-97)
Novices working on the system add the following minutes at each stage.
15, 7, 30, 20, 10, 10.
So any combination that allows us to add 23 minutes at most will satisfy the condition.
EEEEEE (0 mins), NEEEEE(15 mins), ENEEEE(7mins), EEENEE(20mins), EEEENE(10 mins), EEEEEN (10 mins) all work.
Other than this, NNEEEE(22 mins), ENEENE (17 mins), ENEEEN (17 mins), and EEEENN (20 mins) also fit.
Total cases - 10
10/64 = 1.56, approx 1.6
MG
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Tough one... thought for a while.
Max processing time = 120 mins.
Note the following deltas between the expert timing and novice timing (calculated as expert timing -novice timing) :
d1 = +15, d2 = +7, d3= +30, d4 = +20, d5= +10, d6= +10
Sum of times of all experts = 98
Excess time = 120-98 = 22 mins.
We increase the time of processing up to a max of 22 mins.
P = {e1, e2, e3, e4, e5, e6} .............. (1)
Start with all experts, and look at the delta's to replace the experts with the novices, to increase the processing time, upto a max of 22 mins.
d1 = +15, d2 = +7, d3= +30, d4 = +20, d5= +10, d6= +10
a. {n1, e2-e6} (n1=>e1 i.e. n1 replaced e1)ExcessTime: 7mins
b. {n1, n2, e3-e6} (n1=>e1, n2=> e2) ET: 0mins
c. {e1-e3, n4, e5, e6} {n4=>e4} ET: 2 mins
d. {e1, n2, e3, e4, n5, e6} {n2=>e2, n5=> e5} ET: 5 mins
e. {e1, n2, e3, e4, e5, n6} {n2=> e2, n6=>e6} ET: 5 mins
f. {e1-e4, n5, n6} {n5=>e5, n6=>e6} ET: 2 mins
g. {e1, n2, e3-e6} {n2=>e2} ET: 15mins
h. {e1-e4, n5, e6} {n5=>e5} ET: 12mins
i. {e1-e5, n6} {n6=> e6} ET : 12 mins
j. {e1-e6} ET: 22 mins <<< Thanks to Mikki!! >>>>
Total number of ways: 10 for processing in atmost 120mins
Possible # of ways = 64
Prob = 10/64= 0.16
Answer: A
Last edited by haas_mba07 on 29 Aug 2006, 11:29, edited 1 time in total.
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Thats almost exactly how I did it... think you may have missed out the case in which all stages are completed by experts 
Leads to 10 cases.
Cheers
MG
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Aha!!! 
Thanks mikki!!
mikki0000 wrote: Thats almost exactly how I did it... think you may have missed out the case in which all stages are completed by experts Leads to 10 cases. Cheers MG |
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GMAT Instructor
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Good work OA=0.16 I was beginning to think that nobody liked this question!
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