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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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Here we go:


We can arrange n people in n! ways.

Now consider Lisa and Philip as one person.
So we'll have K-1 Employees to be seated on K chairs. This can we done in (k)P(K-1) * 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 (K-1)} * 2!

Probability = {(K) p (K-1)} * 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]
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]
Expert Reply
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]
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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Re: From a group of J employees, K will be selected, at random, to sit in [#permalink]
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 nPr = JPK
(1) K = 15
JP15
We dont know out of how many employee we have to choose 15
INSUFFICIENT

(2) K = J
JPJ OR KPK
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
15P15 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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Just for completeness, if anyone were to read this:

We have to account for 2 possibilities in the Favorable Outcomes:

(1) L-P

Lisa is to the left of Phillip in each arrangement

(2) P-L

Phillip is to the left of Lisa in reach arrangement

Thus,

# of Favorable Outcomes in which Lisa and Phillip sit next to each other is: (2) * (14!)

And

Total Possible Outcomes with no constraints = 15!

Probability = (2 * 14!) / (15!) = (2 * 14!) / (15 * 14!) =

2/15

Not 1/15

ynaikavde wrote:
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= 1-1/15. hence statement 2 with 1 is sufficient ans>> 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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