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A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]
23 Jan 2007, 11:42

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

82% (02:30) correct
18% (00:50) wrong based on 28 sessions

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

The way I am doing is the 1-P(both blue)
P(both blue)=2/8*1/7=1/28
hence probability of both not blue is 1-1/28= 27/28 but the Answer is 15/28. Can someone please explain me what am I doing wrong here?

if the answer is 15/28 the question probably asks for the probability that neither of the two chosen cards is blue.

this is indeed: 6/8*5/7 = 15/28

Yes the answer is 15/28 and I think what you did is right and looks to me another way of doing it, but why is the approach 1-p(both blue) is giving wrong result??

I can see two approaches to solve this:
1) The probability (not choosing blue) = 1- P(both blue)
or
2) just exclude 2 blue cards and choose from the rest 6 cards which is what you did.

Still I do not understand why the first approach is not working??

there is a difference between "neither is blue" and "they are not both blue".

what you did is "not both blue" which, for example, includes the case that one is blue and the other is not. "neither is blue" means that none of the chosen cards is blue".

there is a difference between "neither is blue" and "they are not both blue".

what you did is "not both blue" which, for example, includes the case that one is blue and the other is not. "neither is blue" means that none of the chosen cards is blue".

Okay I see the approach now. However, Going back the same problem solving approach, so can I express this problem as

P(neither is blue) = 1 - p(either is blue)

however I am not sure how to find p(either is blue)?

is it 8C2 total ways = 28, to choose 2 cards of which one is always blue is 2C1*6C1 + 2C2 (when both are blue) = 12 + 1 = 13 hence p(either is blue) = 13/28

and P(neither is blue) = 1 - 13/28 = 15/28

Is the above way right now (may not be preferable cz of complexity though)?

however - in many cases, it is easier to solve "neither" question than "either" question. in fact, many "either" or "at least" questions are better solved as "neither" question, then use the "1-p" rule to get the "either" part.

Re: Probability question [#permalink]
08 Jan 2008, 08:39

paam0101 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

The way I am doing is the 1-P(both blue) P(both blue)=2/8*1/7=1/28 hence probability of both not blue is 1-1/28= 27/28 but the Answer is 15/28. Can someone please explain me what am I doing wrong here?

1st draw: 6/8 will not be blue take one out. 2nd draw: 5/7 will not be blue

6/8 * 5/7 = 3/4*5/7 = 15/28 _________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: Probability question [#permalink]
09 Jan 2008, 23:52

[quote="paam0101"]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

quote]

8 cards 2 blue. Both Blue. 2/8*1/7 = 1/28. Should be 27/28. I think there is a typo here.

Lets try neither is blue. 6/8*5/7 --> 30/56 --> 15/28

Re: Probability question [#permalink]
10 Jan 2008, 00:37

automan wrote:

GMATBLACKBELT wrote:

paam0101 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

quote]

8 cards 2 blue. Both Blue. 2/8*1/7 = 1/28. Should be 27/28. I think there is a typo here.

Lets try neither is blue. 6/8*5/7 --> 30/56 --> 15/28

I get A

There is no type here. Answer is 30/56=15/28

1st approach: P(one blue and the other not blue)=24/56; P(both blue)=2/56; P(both are not blue)=1-26/56=30/56

2nd approach: P(one blue and the other not blue)=24/56; P(both blue)=2/56; P(both are not blue)=(6/8)*(5/8)=30/56

Re: Probability question [#permalink]
26 Aug 2008, 08:21

paam0101 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

The way I am doing is the 1-P(both blue) P(both blue)=2/8*1/7=1/28 hence probability of both not blue is 1-1/28= 27/28 but the Answer is 15/28. Can someone please explain me what am I doing wrong here?

p = 1- (both are blue+ one of them is blue)/total possibilities. = 1- (2C2 +2C1*6C1)/8C2= 1- 13/28 = 15/28 _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: Probability question [#permalink]
12 May 2010, 04:10

automan wrote:

GMATBLACKBELT wrote:

paam0101 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

quote]

8 cards 2 blue. Both Blue. 2/8*1/7 = 1/28. Should be 27/28. I think there is a typo here.

Lets try neither is blue. 6/8*5/7 --> 30/56 --> 15/28

I get A

There is no type here. Answer is 30/56=15/28

1st approach: P(one blue and the other not blue)=24/56; P(both blue)=2/56; P(both are not blue)=1-26/56=30/56

2nd approach: P(one blue and the other not blue)=24/56; P(both blue)=2/56; P(both are not blue)=(6/8)*(5/8)=30/56[/quote]

I have a question why do we exclude situation when one card is blue and other is red, green or yellow? we should exclude just when they both are blue

Re: A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]
14 Sep 2014, 01:08

paam0101 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

The way I am doing is the 1-)P(both blue P(both blue)=2/8*1/7=1/28 hence probability of both not blue is 1-1/28= 27/28 but the Answer is 15/28. Can someone please explain me what am I doing wrong here?

The red highlighted one is the error !!!

i.e For none of the ball to be blue ...you must subtract the probability of getting at least 1 ball blue ( 1-P(atleast 1 blue))

therefore it twill be ..Total no. of ways : 28 < By getting number of ways, you can eliminate C/D/E )

then ways getting at least 1 blue is : 6C1X2C1 + 2C2

Re: A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]
14 Sep 2014, 15:17

Expert's post

paam0101 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

The way I am doing is the 1-P(both blue) P(both blue)=2/8*1/7=1/28 hence probability of both not blue is 1-1/28= 27/28 but the Answer is 15/28. Can someone please explain me what am I doing wrong here?

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. _________________