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

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Manager
Joined: 27 Aug 2004
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A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]  10 Sep 2004, 12:10
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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
Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
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Ok, I think I am doing something totally wrong.

I assume that the total number of cards is 8.

2 blue
2 red
2 yellow
2 green

Otherwise I don't know how many cards contains in total (52?)

What I did:

2/8 * 1/7 = 2/56

Propability both not blue = 56/56 - 2/56 = 54/56 = 27/28

But this is not mentioned in the answers.

Somebody has to help me, what I am doing wrong.

Regards,

Alex
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Joined: 15 Dec 2003
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Why don't you just do the reverse?
Prob of not getting a blue card on first draw: 6/8
Prob of not getting a blue card on second draw: 5/7
Prob of not get blue card when drawing 2 cards: 6/8 * 5/7 = 15/28
_________________

Best Regards,

Paul

Manager
Joined: 02 Apr 2004
Posts: 224
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Kudos [?]: 13 [0], given: 0

Paul,

Thanks a lot! I get your point and idea.
However the other way (not the reverse way) should give the same answer, right?

Why can't I get that?

Regards,

Alex
Joined: 31 Dec 1969
Location: India
Concentration: General Management, Strategy
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
GPA: 3.6
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Kudos [?]: 89 [0], given: 87156

It may b easy to do it this way..

Total outcomes = 8C2.
Two cards are not blue - means other cards = 6C2

6C2/8C2 = 15/28
Director
Joined: 16 Jun 2004
Posts: 893
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Sorry the guest was me..

It may b easy to do it this way..

Total outcomes = 8C2.
Two cards are not blue - means other cards = 6C2

6C2/8C2 = 15/28
Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
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Kudos [?]: 13 [0], given: 0

sorry guys to bother you.

I think I have figured out already.
Question is asking for both not blue.

I forgot the possibility of one still being blue.

Is that correct?

Regards,

Alex
Director
Joined: 08 Jul 2004
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Prob of first non blue 6/8. second. 5/7. Multiply to get 15/28.
A for me.
S
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Kudos [?]: 217 [0], given: 0

Alex_NL wrote:
Paul,

Thanks a lot! I get your point and idea.
However the other way (not the reverse way) should give the same answer, right?

Why can't I get that?

Regards,

Alex

The reason why your method is wrong is because you did not consider the probability when you draw only 1 blue out of the 2 cards
In order to find the probability of not having any blue card, you have to also consider the probability when the first or second card drawn is blue. Hence, you get:

1 - P(2 blue) - P(first blue, second not blue) - P(first not blue, second blue)
1 - (2/8 * 1/7) - (2/8 * 6/7) - (6/8 * 2/7)
1 - 2/28 - 6/28 - 6/28
1 - 13/28 = 15/28
_________________

Best Regards,

Paul

Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
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Kudos [?]: 13 [0], given: 0

I always appreciate help!

Thanks guys. I will study this!

Regards,

Alex
Manager
Joined: 27 Aug 2004
Posts: 126
Location: US
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Kudos [?]: 3 [0], given: 0

I made the same mistake as Alex.
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