R. E. Question

Question

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vivek123 · Answer

If 15 are drawn from 3015 & not a single should be one of the 15 cards added...the probability is:

= (3000 C 15)/(3015 C 15)


= (Number of ways 15 cards can be drawn from original 3000)/(Total number of ways of drawing 15 cards from 3015)

chriswil2005 · Answer

Why is it not, (3000/3015)*(2999/3014)*(2998/3013)....(2985/3000)? Vivek, could you explain your answer a little more. Thanks

sudhagar · Answer

Could both of you explain the answer? 

Macedon: Once answer is explained could you pls post OA?

automan · Answer

In the first attempt the probablity of not taking one of those 15 cards is 3000/3015. In the second attempt is 2999/3014. And so on. In the 15ht attempt the probability is 29985/3000. Multiply all these factors and you will get the answer:

3000*2999*2998*......*2985
----------------------------------
3015*3014*3013*......*3000

olorin · Answer

That's a mighty tough question Macedon to have come up on your GMAT exam yesterday. I think this was an "experimental" question - or I too am hopelessly outclassed by these questions. 

I can't even begin to imagine what the answer choices were.

nakib77 · Answer

I got 1 - (15!)/(315*314*313*312........300)

it is a mighty lengthy one.

sudhagar · Answer

Hope these questions are 'experimental'.

Thank automan for the explanation. However, the answer involves complex number calculation, which GMAT might not expect. Hence lets wait for Macdeon to post the possible answer choices or the format of answer choices (like nCm)

sudhagar · Answer

Let me take a crack at the modified problem:

Total = 3000

prob of not picking = 1 - prob of picking

= 1 - ( 15/3000 * 14/2999 * ... so on). (Note: without repetition). Since this is gonna take a while to compute and given the answer choices are simple numbers. I assume this would be with repetition. This assumption yields:

= 1 - (15/3000)
= 1 - (1/200)
= 199/200

Does this make sense? or Did I fell for any trap?

FN · Answer

this is actually quite simple...more like 650 level question'


Prob of not picking (the 15 cards)=1-(probability of picking that 15 cards)

1-15/3000=1-1/200=199/200

nakib77 · Answer

the question doesn't say he will be drawing 15 cards at the same time. It says he will be drawing it with out repetaion so it go to be 

1- ( 15!/(300*299*298*297.......286)

vivek123 · Answer

Ok, the way I did it is like this...


1) there are 3000 original cards, & 15 new cards are added. So total number of cards 3015.

2) We need to draw 15 cards (without replacements)

3) Total number of ways we can draw 15 cards from the total 3015 cards = 3015C15

4) Total number of ways in which 15 cards can be drawn from original 3000 cards (OR this is same as not drawing from new 15) = 3000C15

=> probability = (3000C15)/(3015C15)


Can anybody explain if there is any problem with this approach???

HongHu · Answer

Total outcome of drawing 15 cards from 3000 cards without replacement would be C(3000, 15). Outcome of drawing 15 cards from (3000-15)=2985 cards would be C(2985, 15). So the odds would be C(2985,15)/C(3000, 15). GMAT would not ask you to get the actual answer.

GMATT73 · Answer

If the question were phrased: 15 cards are added to 2885 to make a total of 3000, and a card is drawn at random without removal, then what is the probability of NOT drawing one of the 15 cards?

Then it becomes a simple probability question 1-15/3000= 1-1/200 = 199/200. This seems more like a realistic ETS 550-600 level question.

Macedon> I am really amazed at how well you can remember so many questions.

HongHu · Answer

Just a reminder we are not supposed to talk about real GMAT questions.

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