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From a group of J employees, K will be selected, at random, to sit in [#permalink]
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18 Feb 2015, 03:55
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Re: From a group of J employees, K will be selected, at random, to sit in [#permalink]
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18 Feb 2015, 06:53
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Selecting k people from J and then arranging then in K seats
Statement1 : No information on J. insufficient not sure if we both will be selected. Statement2 : As we are selecting all then we are just arranging in K seats
If we combine two employees and then we arrange 14 people in 14 seats. this gives us ways when both of the employees sit together. % of time that happens is divided by total possibilities( 15 people in 15 seats). so probability that they sit together can be calculate=14!/15! =1/15 probability that they wont sit together= 11/15. hence statement 2 with 1 is sufficient ans>> C



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From a group of J employees, K will be selected, at random, to sit in [#permalink]
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18 Feb 2015, 08:02
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Here we go:
We can arrange n people in n! ways.
Now consider Lisa and Philip as one person. So we'll have K1 Employees to be seated on K chairs. This can we done in (k)P(K1) * 2! (assuming that we have picked Lisa and Philip in K employees out of J)
Lets proceed:
St1: K = 15
No Information about J
Not sufficient
St2: K = J
Total outcomes = J! or K! Favorable Outcome = {(K) p (K1)} * 2!
Probability = {(K) p (K1)} * 2! / K! > 2 / K
We don't any information about K in St2
not sufficient
Combining:
K = 15
Probability = 2/15
Hence option C is correct.



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Re: From a group of J employees, K will be selected, at random, to sit in [#permalink]
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22 Feb 2015, 09:57
hi all probably my thinking is skewed here, but it is clearly stated in the stimulus that that there are absolutely no restrictions or order of arrangement..isn't that a clue ? aren't we jumping the gun if we calculate the probablity using the permutation formula? i have a feeling that common decision tree analysis should solve the problem..those numbers are just to throw us off. please reveal the official solution and explanation asap. will be a great help if a math expert out there in this delightful forum would care to throw some light onto this



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Re: From a group of J employees, K will be selected, at random, to sit in [#permalink]
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22 Feb 2015, 11:05
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Bunuel wrote: From a group of J employees, K will be selected, at random, to sit in a line of K chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Lisa is seated exactly next to employee Phillip?
(1) K = 15 (2) K = J
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTIONStatement #1: given this, we know K = 15 employees are seated, but we have no idea what the size of the larger pool is. If J is much larger than K, then it becomes unlikely that either Lisa or Phillip is among those seated. Both would have to be among the seated in order for them to have a chance of sitting next to each other. Even if J is closer to K, then it becomes more likely that both are among the seated, but we would need to know the number for J to complete the calculation. This statement, alone and by itself, is insufficient. Statement #2: Ignore the information in the first statement. If M = N, then all employees, everyone in the pool of selection, takes a seat. The problem is — if M = N is a small number, say 4 or 5, then it would be more likely that Lisa and Phillip would wind up next to each other, but if M = N is large, say 200, then its quietly likely that two specific employees wind up nowhere near each other. The size of M = N would make a big different in the calculation, and without knowing that, we can’t calculate. This statement, alone and by itself, is insufficient. Combined statements: Now, M = N = 15. We are seating a group of 15 employees, and we want to know whether two of the people seated, Lisa and Andrew, are next to each other. This is a calculation we can perform, using counting techniques. With this combined information, we can give a definitive numerical answer to the prompt question. Together, the statements are sufficient. Statements sufficient together but no individually. Answer = C  See more at: http://magoosh.com/gmat/2013/gmatdata ... JzRXH.dpuf
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Re: From a group of J employees, K will be selected, at random, to sit in [#permalink]
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22 Feb 2015, 13:41
Hi smashbiker84, A certain number of DS questions on Test Day can be "talked through", without doing much math at all  this prompt is an example of that concept. The tricky part on Test Day is in finding the 'balance' between doing work and just trying to think your way through the logic. DS questions have NO 'safety net', which means that if you make a little mistake (or you miss one of the possible answers), then you'll likely get the question wrong AND you won't even know it (or be able to catch the error). As such, you should NOT be attempting to talk your way past many DS questions. In most cases, the work is remarkably easy, so if your goal is to score at a high level, then you have to be willing to do what it takes to guarantee that score result  write everything down, do the work on the pad and PROVE that you're correct. GMAT assassins aren't born, they're made, Rich
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Re: From a group of J employees, K will be selected, at random, to sit in [#permalink]
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05 May 2015, 19:04
a) We don't know whether Lisa and Phillip selected from the group. So not sufficient to calculate the probability. A and D are not answers. b) IF J=K even then we cannot calculate the total number cases due to limitation of not knowing any value. Not sufficient. B is not an answer. A+B=C) We know the value of K=15 from statement 1 and J=K=15 the size of group. We can easily calculate the total cases and favorable cases but we don't need because it DS(:P). Answer is C.



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From a group of J employees, K will be selected, at random, to sit in [#permalink]
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04 Aug 2016, 04:51
Bunuel wrote: From a group of J employees, K will be selected, at random, to sit in a line of K chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Lisa is seated exactly next to employee Phillip?
(1) K = 15 (2) K = J
Kudos for a correct solution. From Stimulus n Pr = J PK (1) K = 15 J P15 We dont know out of how many employee we have to choose 15 INSUFFICIENT (2) K = J J PJ OR K PK We dont know either K or J All we know is this expression = K! (or J! since K and J are same ) INSUFFICIENT MERGING BOTH 15 P15 Which is equal to 15!Now even though it is a time consuming task ,we can calculate all permutations when LISA and PHILIP are sitting together SUFFICIENT ANSWER IS C
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Re: From a group of J employees, K will be selected, at random, to sit in [#permalink]
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