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A certain deck of cards contains 2 blue cards, 2 red cards,

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A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink] New post 19 Jul 2004, 11:31
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A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?
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 [#permalink] New post 19 Jul 2004, 12:00
1 - P(they both will be blue) =

1 - 2/8x1/7=

1 - 1/28 = 27/28

is that right?
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 [#permalink] New post 19 Jul 2004, 12:04
Nope! This is actually was my answer too, but OA is different :?
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 [#permalink] New post 19 Jul 2004, 13:33
I got the same answer as Ian, 27/28.

It'd be interesting to hear the OA and its explanation.
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 [#permalink] New post 19 Jul 2004, 13:41
Let's give the chance others first. Guys! Any ideas!! :!: :roll:
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 [#permalink] New post 19 Jul 2004, 22:52
I approached in 2 possible way (which I know..) but I got the same answer as others got.

Approach 1 :
Total Possible ways = 8C2 = 28
When both cards are blue = 2C2 = 1

P (when both blue) = 1/28
P ( when both are not blue) = 1 - 1/28 = 27/28

Approach 2:
Drawing 2 balls when none is blue = 6C2 = 15
Drawing 2 balls when 1 is blue = 6C1 * 2C1 = 12

Total = 27

Required Probability = 27/28


Please let me know which step I have overlooked... (in both approach..)
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Re: PS cards [#permalink] New post 19 Jul 2004, 23:01
boksana wrote:
A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?


The possibility that the first one isn't blue = 6/8 = 3/4
The possibility that the second one isn't blue is = 5/7

--->the probability that they will both are not blue = 3/4*5/7 = 15/28
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 [#permalink] New post 19 Jul 2004, 23:49
But your case assumes that the ball was not replaced....

Or may be that I have assumed that ball was replaced..


Which one is right Boksana ??
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 [#permalink] New post 20 Jul 2004, 05:21
i think we are overlookin something here

p(first card is not blue) = 7/8
p(second card is not blue) = 6/7

p( two cards are not blue)=(7/8)*(6/7)=3/4

Is this true?
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 [#permalink] New post 20 Jul 2004, 06:24
IMO, answer should be 7/8

Probability that two cards will be blue = 2C1/2^4 = 1/8
Not Blue = 1-1/8=7/8
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 [#permalink] New post 20 Jul 2004, 07:49
the answer, 15/28, answers the question: "what is the probability that neither card is blue?"

This isn't a trick question, it's just not worded right for the answer desired. The question says, "what is the probability that both will not be blue"? In probability speak, that means together. It doesn't eliminate the possibilty of having a blue card with a different color card.

Again, the GMAT will not be this ambiguous. If we can sit here and argue about a question, it won't be on the test. The GMAT doesn't ever want to leave open the possibility that someone would contest a question. They make them rock solid, even if they're confusing.

That said, many books on the market have no such policy. So if you get a princeton review book, a kaplan book, an arco book, or any other book that someone is publishing once a year to try to cash in on the GMAT craze, where the ramifications of publishing a bad question aren't so heavy, expect questions with ambiguous solutions.
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 [#permalink] New post 20 Jul 2004, 08:08
This was the question from Delta course. I've got 27/28 as well and wanted to see how you guys would solve it.
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 [#permalink] New post 21 Jul 2004, 12:58
I totally agree with Ian that "...both will not be blue" in probability speaking means together.

The ans 15/28 says that the question should be sound like "what is the probability that neither card is blue?"

Prob (neither card is blue) = 1 - Prob (at least one is blue)

Prob (at least one is blue) = 2/8 x 6/7 x 2 + 2/8 x 1/7 = 13/28

Prob (neither card is blue) = 15/28

My regards
  [#permalink] 21 Jul 2004, 12:58
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