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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 10 Sep 2004, 12:10
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A
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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
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 [#permalink] New post 10 Sep 2004, 12:23
Ok, I think I am doing something totally wrong.

I get an different answer.

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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 [#permalink] New post 10 Sep 2004, 12:33
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
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Paul

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 [#permalink] New post 10 Sep 2004, 12:38
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
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 [#permalink] New post 10 Sep 2004, 12:45
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
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 [#permalink] New post 10 Sep 2004, 12:47
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
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 [#permalink] New post 10 Sep 2004, 12:48
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
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 [#permalink] New post 10 Sep 2004, 12:49
Prob of first non blue 6/8. second. 5/7. Multiply to get 15/28.
A for me.
S
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 [#permalink] New post 10 Sep 2004, 13:06
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
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Paul

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 [#permalink] New post 10 Sep 2004, 13:13
:good :good :good

I always appreciate help!

Thanks guys. I will study this!

Regards,

Alex
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 [#permalink] New post 10 Sep 2004, 17:48
A is the answer.
I made the same mistake as Alex.
  [#permalink] 10 Sep 2004, 17:48
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A certain deck of cards contains 2 blue cards, 2 red cards,

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