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

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Manager
Joined: 18 Oct 2010
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A certain deck of cards contains 2 blue cards, 2 red cards [#permalink]

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05 May 2012, 03:08
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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
Manager
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink]

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05 May 2012, 04:04
8
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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]

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05 May 2012, 03:26
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Expert's post
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.

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]

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07 May 2012, 16:43
3
KUDOS
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
Intern
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink]

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11 May 2012, 18:41
2
KUDOS
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
Manager
Joined: 18 Oct 2010
Posts: 79
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink]

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11 May 2012, 23:53
1
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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
Manager
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Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink]

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07 May 2012, 18:59
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?
Intern
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Kudos [?]: 0 [0], given: 3

Re: A certain deck of cards contains 2 blue cards, 2 red cards [#permalink]

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20 Oct 2012, 22:01
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

Manager
Joined: 15 Aug 2012
Posts: 111
Location: India
Concentration: Technology, Strategy
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A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]

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13 Nov 2012, 19:36
Probability that both are not blue means both are other colors(except blue)

6C2 / 8C2 = 15/28
Intern
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GMAT Date: 01-16-2013
GPA: 3.37
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Kudos [?]: 2 [0], given: 6

Re: A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]

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13 Nov 2012, 23:15
=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
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A certain deck of cards contains 2 blue [#permalink]

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02 Nov 2015, 04:51
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

This is a post from one of the earlier posts in the forum. All the replies assumed that the cards are drawn one by one, but this assumption is not mentioned in the question.
Why cannot we draw 2 cards together and get the answer as 6/8.
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Re: A certain deck of cards contains 2 blue [#permalink]

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02 Nov 2015, 05:03
sd.1223 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?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

This is a post from one of the earlier posts in the forum. All the replies assumed that the cards are drawn one by one, but this assumption is not mentioned in the question.
Why cannot we draw 2 cards together and get the answer as 6/8.

Hi sd.1223, I also had the same problem

However, the default assumption is that cards are always drawn one by one unless there is clear indication that they will be drawn together / drawn with replacement.
Re: A certain deck of cards contains 2 blue   [#permalink] 02 Nov 2015, 05:03
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