Probability/counting - select 5 numbers ...

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?

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.

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..

Refer the attachment for more information on probability distribution.

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.

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.

Hardik,
the material is excellent.
but binomial and poisson dist are not in scope of gmat i guess

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.

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.

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..

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..