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

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Senior Manager
Joined: 09 Aug 2005
Posts: 283
A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]

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22 Feb 2006, 22:05
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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?
A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

scroll for the oa which i think is wrong can you please confirm ?

OA - A -
OE 1 st card that is not blue = 6/8
2nd card that is not blue = 5/ 7

= 30/56 = 15/ 28 - how do you conclude that the cards are draw one after the other?

Another method: 1 - P(at least 1 blue card)
but what about p(both not blue)

I think it should be plain 1- (1/28) = 27/28
VP
Joined: 29 Dec 2005
Posts: 1341
Re: PS - probability cards [#permalink]

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22 Feb 2006, 22:14
= 1-p(both blue)
=1-1/8c2
=1-1/28
=27/28

but you would be far better with question posting without OA. OA is not the problem. we post question here to get different approaches and viewpoints. posting OA blocks lowers the chance of getting responses.
SVP
Joined: 14 Dec 2004
Posts: 1689

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24 Feb 2006, 22:08
1-(1/28)
= 27/28.

CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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24 Feb 2006, 23:11
I think OA is correct.

Total ways = 8P2 = 56

Ways when atleats oen of them is blue

BB - 2 cases (B1B2, B2B1)
BR - 4 cases (B1R1, B2R1, B1R2, B2R2)
RB - 4 cases
BY - 4 cases
YB - 4 cases
BG - 4 cases
GB - 4 cases

Total when atleast 1 is blue = 26

Total when none of them is blue = 30

Prob = 30/56 = 15/28

I think you forgot to take into account when one of them is blue. Question would be more clear if this would have been written "None of them is blue."
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

24 Feb 2006, 23:11
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