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Bill has a small deck of 12 playing cards made up of only 2 [#permalink]
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21 Jan 2007, 17:16
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68% (03:05) correct
32% (02:43) wrong based on 157 sessions
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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: billhasasmalldeckof12playingcardsmadeupofonly2suitsof6cardseach96078.html
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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: 116/33 = 17/33



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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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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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Thanks a lot pal!
Probability sure is one of my weakpoints......



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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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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



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C.
p=1q, where q is the probability to turn over 4 cards with different values.
12P4  the total number of combinations.
12*(111)*(102)*(93)=12*10*8*6  the number of combinations with 4 cards with different values.
p=112*10*8*6/12P4=112*10*8*6*8!/12!=112*10*8*6/(12*11*10*9)=18*6/(11*9)=116/33=17/33



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Combinations [#permalink]
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15 Dec 2007, 19:44
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]
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15 Dec 2007, 23:09
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 > 116/33 = 17/33.



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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]
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16 Dec 2007, 07:47
had the same problem with this question: tried to use combinatorics, but failed
can someone explain how to attack this question with comb.tools?



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Re: Combinations [#permalink]
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16 Dec 2007, 08:20
elgo wrote: had the same problem with this question: tried to use combinatorics, but failed 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 more2minutes method.



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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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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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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+(10C21)+(10C22)+(10C23)+(10C24)+(10C25) ]/12C4=17/33
Now i can rest my mind



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Re: Combinations [#permalink]
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16 Dec 2007, 10:09
walker wrote: elgo wrote: had the same problem with this question: tried to use combinatorics, but failed can someone explain how to attack this question with comb.tools? see this one: http://www.gmatclub.com/forum/t56530I guess If we have a problem with complex restrictions, nCm, nPm formulas will be too unwieldy and a more2minutes 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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walker wrote: p=1q, 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
(1211)/11 * (1222)/10 * (1233)/9 = 16/33
1  16/33 = 17/33



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Re: cards [#permalink]
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22 Dec 2007, 10:37
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 > 116/33 =17/33
Ok there we go.



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Re: cards [#permalink]
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05 Jan 2008, 09:19
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 > 116/33 =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.....







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