Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 16 Jan 2007
Posts: 18

A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]
Show Tags
23 Jan 2007, 12:42
2
This post was BOOKMARKED
Question Stats:
82% (02:18) correct
18% (01:48) wrong based on 130 sessions
HideShow timer Statistics
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 1P(both blue)
P(both blue)=2/8*1/7=1/28
hence probability of both not blue is 11/28= 27/28 but the Answer is 15/28. Can someone please explain me what am I doing wrong here?



Senior Manager
Joined: 23 Jun 2006
Posts: 387

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



Intern
Joined: 16 Jan 2007
Posts: 18

hobbit wrote: 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 1p(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??



Senior Manager
Joined: 23 Jun 2006
Posts: 387

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".



Intern
Joined: 16 Jan 2007
Posts: 18

hobbit wrote: 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)?



Senior Manager
Joined: 23 Jun 2006
Posts: 387

your way and approach are now correct.
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 "1p" rule to get the "either" part.
i'll give a nice example in a new thread...



CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

Re: Probability question [#permalink]
Show Tags
08 Jan 2008, 09:39
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 1P(both blue) P(both blue)=2/8*1/7=1/28 hence probability of both not blue is 11/28= 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



CEO
Joined: 29 Mar 2007
Posts: 2559

Re: Probability question [#permalink]
Show Tags
10 Jan 2008, 00:52
[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
I get A



Director
Joined: 09 Jul 2005
Posts: 591

Re: Probability question [#permalink]
Show Tags
10 Jan 2008, 01:37
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)=126/56=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



SVP
Joined: 07 Nov 2007
Posts: 1799
Location: New York

Re: Probability question [#permalink]
Show Tags
26 Aug 2008, 09:21
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 1P(both blue) P(both blue)=2/8*1/7=1/28 hence probability of both not blue is 11/28= 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



Intern
Joined: 21 Feb 2010
Posts: 33
Location: Ukraine

Re: Probability question [#permalink]
Show Tags
12 May 2010, 05:10
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)=126/56=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



Manager
Joined: 16 Feb 2010
Posts: 186

Re: Probability question [#permalink]
Show Tags
14 May 2010, 14:25
but the question is vague, nowhere is mentioned the it is with repalcement or without replacement.



Manager
Joined: 16 Feb 2010
Posts: 186

Re: Probability question [#permalink]
Show Tags
17 May 2010, 11:06
i misread the question sorry!!



Manager
Joined: 28 Aug 2013
Posts: 98
Location: India
Concentration: Operations, Marketing
GMAT Date: 08282014
GPA: 3.86
WE: Supply Chain Management (Manufacturing)

Re: A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]
Show Tags
14 Sep 2014, 02:08
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 11/28= 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 ( 1P(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 you will get A : Regards LS
_________________
Gprep1 540 > Kaplan 580>Veritas 640>MGMAT 590 >MGMAT 2 640 > MGMAT 3 640 > MGMAT 4 650 >MGMAT 5 680  >GMAT prep 1 570
Give your best shot...rest leave upto Mahadev, he is the extractor of all negativity in the world !!



Math Expert
Joined: 02 Sep 2009
Posts: 39704

Re: A certain deck of cards contains 2 blue cards, 2 red cards, [#permalink]
Show Tags
14 Sep 2014, 16:17
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 1P(both blue) P(both blue)=2/8*1/7=1/28 hence probability of both not blue is 11/28= 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=1P(blue, blue)=12/8*1/7=27/28.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Re: A certain deck of cards contains 2 blue cards, 2 red cards,
[#permalink]
14 Sep 2014, 16:17








Similar topics 
Author 
Replies 
Last post 
Similar Topics:




If 2 cards are selected at random from the deck of 52 cards then What

GMATinsight 
1 
27 Jul 2015, 05:53 

7


There are 2 decks of cards. The first deck has a 100 cards labelled...

kdatt1991 
5 
23 Jul 2016, 08:43 

21


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

Joy111 
11 
02 Nov 2015, 05:03 



A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar

rxs0005 
8 
01 Sep 2010, 23:51 

4


A deck of 9 cards contains 2 red cards, 3 blue cards, and 4 green card

gmatcracker2010 
2 
29 Mar 2016, 13:46 



