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Probability/counting - select 5 numbers ...

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by Brent@GMATPrepNow » Thu Jan 29, 2009 8:50 am
If five numbers are randomly selected without replacement from the set {1,2,3,4,5,6,7,8,9}, what is the probability that, of the five selected numbers, the second largest is 6?
A) 5/126
B) 5/112
C) 5/56
D) 5/21
E) 5/14

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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by hardik.jadeja » Thu Jan 29, 2009 10:32 am
IMO answer is D 5/21.

Its a hypergeometric probability distribution.

Assume we have 3 sets of number here.
Set 1: numbers greater than 6. {7,8,9}
Set 2: It contains only number 6. {6}
Set 3: numbers less than 6. {1,2,3,4,5}

We have to select 1 number from set1, one from set 2 and 3 numbers from set 3. This way 6 will be the second largest number among the chosen 5 numbers.

Probability is (3C1*1C1*5C3) / 9C5, which is 5/21.

Whats OA Brent?
Last edited by hardik.jadeja on Fri Jan 30, 2009 6:41 am, edited 1 time in total.
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by Brent@GMATPrepNow » Thu Jan 29, 2009 10:41 am
5/21 it is - nice work!
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by masuarezdl » Thu Jan 29, 2009 10:52 am
Hardik, are you sure? It is not the same probability to pick a number from a set of 3 than from a set of 8.
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by hardik.jadeja » Thu Jan 29, 2009 11:03 am
I think 5/21 is the correct answer.

You can think of this solution in a diff way also.

Assume that an urn contains 9 balls, 5 red, 1 green, and 3 blue. You have to obtain 3 red balls, 1 green ball and 1 blue ball in drawing five balls without replacement.

Now apply hypergeometric probability distribution.

Blue balls here represent set 1 {7,8,9}
Green ball represent set 2 {6}
and Red balls represent set 3 {1,2,3,4,5}.

HTH..
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by hardik.jadeja » Thu Jan 29, 2009 11:23 am
Refer the attachment for more information on probability distribution.
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PROBABILITY DISTRIBUTIONS.doc
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by 720dreaming » Thu Jan 29, 2009 5:16 pm
Can somebody explain why the following method does not produce the correct answer?

Largest number must be 7, 8 or 9 so 1/3 prob.

Second largest must be 6. 1/8 prob.

Third must be 1, 2, 3, 4, or 5 and 7 are left...so 5/7 prob

The forth would be 4/6 prob

And the last number has to be 3/5 prob

so (1/3)(1/8)(5/7)(4/6)(3/5).

Where is my error?

Thanks.
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by umaa » Thu Jan 29, 2009 7:14 pm
hardik.jadeja, the doc you've attached is too good. very useful. BTW, can you send me or PM the source? I'm too bad at probability.
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by aroon7 » Thu Jan 29, 2009 8:50 pm
Hardik,
the material is excellent.
but binomial and poisson dist are not in scope of gmat i guess
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by masuarezdl » Thu Jan 29, 2009 8:55 pm
hardik.jadeja wrote:Refer the attachment for more information on probability distribution.
Extremely useful mate! Thanks for the answers and the document, it is really helpful. B-)
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by piyush_nitt » Thu Jan 29, 2009 10:36 pm
hardik.jadeja wrote:IMO answer is D 5/21.

Its a hypergeometric probability distribution.

Assume we have 3 sets of number here.
Set 1: numbers greater than 6. {7,8,9}
Set 2: It contains only number 6. {6}
Set 3: numbers less than 6. {1,2,3,4,5}

We have to select 1 number from set1, one from set 2 and 3 numbers from set 3. This way 6 will be the second largest number among the chosen 5 numbers.

Probability is (3C1*1C1*5C3) / 9C3, which is 5/21.

Whats OA Brent?
Denominator should be 9C5 isn't it? or I am missing something here.
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by hardik.jadeja » Fri Jan 30, 2009 6:51 am
Piyush: Yeah.. rightly pointed out.. it was a typo.. thanx.. it should be 9C5.. I have corrected my earlier post..

Aroon: Yes, probably it is out of scope for GMAT but as Brent said this was his attempt to create difficult GMAT-style questions.. and little extra knowledge is never harmful.. :D

Umaa: I found this material online somewhere a long back.. i don't remember the source.. If i find anything useful like this one, i will surely post it in this forum.. :D
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