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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 05 May 2012, 02:08
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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 be not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16


Hi I have a confusion as to how to solve this question , if somebody can help me ?
[Reveal] Spoiler: OA
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink] New post 05 May 2012, 03:04
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Thanks Bunuel for the prompt response.

I quess the these two questions should clear the concept

1)Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that NONE of them are blue?

answer: (4/7) x (3/6) x (2/5) = 4/35


2)Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?


A. 5/7
B. 23/24
C. 6/7
D. 34/35
E. 8/13

so should we work like this : probability of all blue 3/7*2/6*1/6=1/35
none blue : 1-(1/35)= 34/35

we should be clear of the wordings , shouldn't we ?
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink] New post 05 May 2012, 02:26
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BhaskarPaul 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 be not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16

Hi I have a confusion as to how to solve this question , if somebody can help me ?


The question is a bit ambiguous. I guess we are asked to find the probability that neither of the cards drawn is blue, if so then P=P(not blue)*P(not blue)=6/8*5/7=15/28.

Answer: A.

One can interpret the question as: what is the probability that we don't have (blue, blue) case then P=1-P(blue, blue)=1-2/8*1/7=27/28.

P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.

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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink] New post 07 May 2012, 15:43
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Joy111 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 be not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16


Hi I have a confusion as to how to solve this question , if somebody can help me ?

hello

I have computed the probability for both Cards to be blue and substract one
hence total card to 8
probability for the first card 2/8 OR 1/4
For the second 1/7
HENCE 1/28 FOR CARDS

1-1/28 = 27/28

HOPE this helps
best regards
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink] New post 11 May 2012, 17:41
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Joy111 wrote:
keiraria wrote:
Joy111 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 be not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16


Hi I have a confusion as to how to solve this question , if somebody can help me ?

hello

I have computed the probability for both Cards to be blue and substract one
hence total card to 8
probability for the first card 2/8 OR 1/4
For the second 1/7
HENCE 1/28 FOR CARDS

1-1/28 = 27/28

HOPE this helps
best regards



@keiraria

Thanks for your help but can you solve this one for me too, its same as above only the language has been changed , its NONE now .

If it states that both are not blue then one can be blue , try this one below.


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 NONE are blue?


hello Joy
thanks for asking me . in fact the way the question is asked has double meaning and it is
a little ambigious
it is noone are blue it is what I have done :27/28


but if it is that both can not be blue
that is the probability to have noNE blue + the probablity to have maximum one blue and hence one of any other colors

hence or case

27/28+ 2/8x6/7

Hope it is right and make sense to you

best regards
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink] New post 11 May 2012, 22:53
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@ keira

Thanks for your help but can you solve this one for me too, its same as above only the language has been changed , its NONE now .

If it states that both are not blue then one can be blue , try this one below.


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 NONE are blue?[/quote]

hello Joy
thanks for asking me . in fact the way the question is asked has double meaning and it is
a little ambigious
it is noone are blue it is what I have done :27/28


but if it is that both can not be blue
that is the probability to have noNE blue + the probablity to have maximum one blue and hence one of any other colors

hence or case

27/28+ 2/8x6/7

Hope it is right and make sense to you

best regards[/quote]


I quess the these two questions should clear the concept

1)Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that NONE of them are blue?

answer: (4/7) x (3/6) x (2/5) = 4/35


2)Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?


A. 5/7
B. 23/24
C. 6/7
D. 34/35
E. 8/13

so should we work like this : probability of all blue 3/7*2/6*1/6=1/35
none blue : 1-(1/35)= 34/35

for none you do not have to subtract from one
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink] New post 07 May 2012, 17:59
keiraria wrote:
Joy111 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 be not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16


Hi I have a confusion as to how to solve this question , if somebody can help me ?

hello

I have computed the probability for both Cards to be blue and substract one
hence total card to 8
probability for the first card 2/8 OR 1/4
For the second 1/7
HENCE 1/28 FOR CARDS

1-1/28 = 27/28

HOPE this helps
best regards



@keiraria

Thanks for your help but can you solve this one for me too, its same as above only the language has been changed , its NONE now .

If it states that both are not blue then one can be blue , try this one below.


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 NONE are blue?
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink] New post 20 Oct 2012, 21:01
Joy111 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 be not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16


Hi I have a confusion as to how to solve this question , if somebody can help me ?


The question asks " they will both be not blue". Considering that these are draws without replacement there are following possibilities :
draw 1, draw 2
===== =====
1) Not Blue, Blue;
2) Blue, Not Blue;
3) Blue, Blue;
4) Not Blue, not Blue;

The answer has to be either P(4) OR all exclusion of the 3 previous conditions i.e. 1- [P(1) + P(2) + P(3)] => neither of the draws are blue

P(4) = 6/8*5/7=15/28

1- [P(1) + P(2) + P(3)] = 1 - [ (6/8 * 2/7) + (2/8*6/7) + (2/8*1/7)] = 1 - [26/8*7] = 1 - [13/28] = 15/28

both approaches have same answer
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A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink] New post 13 Nov 2012, 18:36
Probability that both are not blue means both are other colors(except blue)

6C2 / 8C2 = 15/28
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Re: A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink] New post 13 Nov 2012, 22:15
Answer = 6C2 / 8C2
=15/28

Total ways of selecting card = 8C2
Favorable = 6C2 (neglecting 2 blue cards)

PS: You did not consider the case of atleast one blue. which can be picked 12 ways.
=1-2C2/8C2-12/8C2
=1-(1/28)-(12/28)
=(28-1-12)/28
=15/28
Re: A certain deck of cards contains 2 blue cards, 2 red cards,   [#permalink] 13 Nov 2012, 22:15
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