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Bill has a small deck of 12 playing cards made up of only 2

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Bill has a small deck of 12 playing cards made up of only 2 [#permalink] New post 21 Jan 2007, 16:16
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Question Stats:

68% (03:17) correct 32% (02:52) wrong based on 85 sessions
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33

OPEN DISCUSSION OF THIS QUESTION IS HERE: bill-has-a-small-deck-of-12-playing-cards-made-up-of-only-2-suits-of-6-cards-each-96078.html
[Reveal] Spoiler: OA

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 [#permalink] New post 21 Jan 2007, 16:27
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answer is C (17/33) in my opinion.
here is how i did it....

total number of 4 cards drawings: 12*11*10*9

drawings which have no pairs at all: 12*10*8*6

probability to have no pairs: 12*10*8*6/12*11*10*9 = 16/33

probability to have at least one pair: 1-16/33 = 17/33
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 [#permalink] New post 22 Jan 2007, 03:54
Hey Hobbit! Could you please explain how you got that figure "Drawings which have no pairs at all"??

I didn't follow that step.
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 [#permalink] New post 22 Jan 2007, 04:40
counting quadruples with no pairs:
first card - no restrictions - 12 options
second card - cannot be same as first - 10 options (out of 11 cards)
third card - cannot be same as first or second - 8 options (out of 10 cards left)
forut card - cannot be same as first three - hence 6 options out of 9 remaining cards).

total 12*10*8*6/12*11*10*9 = 16/33
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 [#permalink] New post 22 Jan 2007, 10:40
Thanks a lot pal! :-D

Probability sure is one of my weak-points...... :(
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 [#permalink] New post 26 Jan 2007, 09:47
Hobbit is right. OA is C.

This problem is from MGMAT. IMO, it's a little too hard to be on the real test but it's good practice nonetheless.
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cards [#permalink] New post 02 Dec 2007, 05:15
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?


A 8/33

B 62/165

C 17/33


D103/165

E 25/33

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 [#permalink] New post 09 Dec 2007, 00:37
Expert's post
C.

p=1-q, where q is the probability to turn over 4 cards with different values.

12P4 - the total number of combinations.

12*(11-1)*(10-2)*(9-3)=12*10*8*6 - the number of combinations with 4 cards with different values.

p=1-12*10*8*6/12P4=1-12*10*8*6*8!/12!=1-12*10*8*6/(12*11*10*9)=1-8*6/(11*9)=1-16/33=17/33
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Combinations [#permalink] New post 15 Dec 2007, 18:44
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

A 8/33
B 62/165
C 17/33
D 103/165
E 25/33

I know the answer is C but i can't understand why i can't get the answer in the following way....
12C4 gives all the combinations = 11*9*5
6*10C2 gives all the possible combinations with at least one pair=6*9*5
divide giving 6/11 ?????
I don't see what i am doing wrong
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Re: Combinations [#permalink] New post 15 Dec 2007, 22:09
alexperi wrote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

A 8/33
B 62/165
C 17/33
D 103/165
E 25/33

I know the answer is C but i can't understand why i can't get the answer in the following way....
12C4 gives all the combinations = 11*9*5
6*10C2 gives all the possible combinations with at least one pair=6*9*5
divide giving 6/11 ?????
I don't see what i am doing wrong


I dunno, bout that approach, im really bad using the combinatorics approach.

Well try and find the probabilty that there are no pairs.
Lets say we pick the following cards: 1,2,3,4. In this order.

1 * 10/11 * 8/10 * 6/9

1: b/c the probability of picking any number is 1.

10/11: is the prob of not picking the next pair
we have 1,2

8/10: we have 1,2,3,3,4,4,5,5,6,6 left. 10 numbers and 2 unfavorables here. So its 8/10

6/9: we have 1,2,3,4,4,5,5,6,6 left. So 9 numbers and 3 unfavorables so 6/9.

We have 16/33 --> 1-16/33 = 17/33.
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 [#permalink] New post 16 Dec 2007, 04:18
Thanks GMATBLACKBELT but i know and understand your approach. I am just trying to standardise my thinking. If combinatorics approach doesn't apply here then i will be confused when to use and when not to use combinatorics.
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Re: Combinations [#permalink] New post 16 Dec 2007, 06:47
had the same problem with this question: tried to use combinatorics, but failed :oops:

can someone explain how to attack this question with comb.tools?
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Re: Combinations [#permalink] New post 16 Dec 2007, 07:20
Expert's post
elgo wrote:
had the same problem with this question: tried to use combinatorics, but failed :oops:

can someone explain how to attack this question with comb.tools?


see this one: http://www.gmatclub.com/forum/t56530

I guess If we have a problem with complex restrictions, nCm, nPm formulas will be too unwieldy and a more-2-minutes method.
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 [#permalink] New post 16 Dec 2007, 07:53
Still do not see what i am missing out or including with the method i used.

"12C4 gives all the combinations = 11*9*5
6*10C2 gives all the possible combinations with at least one pair=6*9*5
divide giving 6/11"
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 [#permalink] New post 16 Dec 2007, 08:08
Expert's post
alexperi wrote:
"12C4 gives all the combinations = 11*9*5

You should use 12P4 instead of 12C4. 1234 and 4321 are different cases.

alexperi wrote:
6*10C2"
Sorry, I don't understand logic of this formula.
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 [#permalink] New post 16 Dec 2007, 08:52
Hmmmmm i think you have struck a cord
6*10C2
There are Six sets if (10C2).
Assume i have selected the first pair then i have i have 10 cards to choose 2 from. And there are 6 such scenarios because there are 6 pairs of cards.

BUT you have just reminded me that this way includes some duplicates so my method nearly correct.

so
[10C2+(10C2-1)+(10C2-2)+(10C2-3)+(10C2-4)+(10C2-5) ]/12C4=17/33

Now i can rest my mind :-D
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Re: Combinations [#permalink] New post 16 Dec 2007, 09:09
walker wrote:
elgo wrote:
had the same problem with this question: tried to use combinatorics, but failed :oops:

can someone explain how to attack this question with comb.tools?


see this one: http://www.gmatclub.com/forum/t56530

I guess If we have a problem with complex restrictions, nCm, nPm formulas will be too unwieldy and a more-2-minutes method.


Thanks a ton!

Anyway, I guess I'd better use probability formula in such cases... nCm, nPm are too much thinking:) (I mean in this particular task)
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 [#permalink] New post 22 Dec 2007, 09:10
walker wrote:

p=1-q, where q is the probability to turn over 4 cards with different values.



Same starting point. But aftwerwards applied the following logic

probability that the second card will not make a pair to the first withdrwan
times
probability that the third will not make a pair with any of the two previously witdrawn cards
times
probability that the fourth one will not make a pair with any of the three previously withdrawn cards

(12-1-1)/11 * (12-2-2)/10 * (12-3-3)/9 = 16/33

1 - 16/33 = 17/33
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Re: cards [#permalink] New post 22 Dec 2007, 09:37
Amardeep Sharma wrote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?


A 8/33

B 62/165

C 17/33


D103/165

E 25/33

Amar


OK here it is, I originally got the answer, but did it through some method that doesn't really make a whole lot of sense to me anymore. I dont think it was correct either, I just got really lucky and somehow things worked out =)

Probability of NOT picking any pairs.

1*10/11* 8/10*6/9

1/11*8*2/3 --> 16/33 --> 1-16/33 =17/33

Ok there we go.
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Re: cards [#permalink] New post 05 Jan 2008, 08:19
GMATBLACKBELT wrote:
Amardeep Sharma wrote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?


A 8/33

B 62/165

C 17/33


D103/165

E 25/33

Amar


OK here it is, I originally got the answer, but did it through some method that doesn't really make a whole lot of sense to me anymore. I dont think it was correct either, I just got really lucky and somehow things worked out =)

Probability of NOT picking any pairs.

1*10/11* 8/10*6/9

1/11*8*2/3 --> 16/33 --> 1-16/33 =17/33

Ok there we go.


can someone tell me where I am wrong????

all possible outcomes: 12c4=990

let's choose 1 pair of identic cards from the 2 sets: 6c1 possibilities=6
let's multiply this result * 10c2, combinations for the 2 free places=6*45=270

let's consider the possibility that we have 2 pairs of identic cards: 6c2=15

thus we have 285/990 = 19/33....please help! I add 15 but I am sure it needs to subtract it from 270.....
Re: cards   [#permalink] 05 Jan 2008, 08:19
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