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Parth
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Post subject: Bill has a small deck of 12 playing cards  Posted: Tue Dec 11, 2007 3:27 pm |
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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?
Can we solve this using Permutation and combination ?
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shaji
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Post subject: Re: Bill has a small deck of 12 playing cards Posted: Thu Dec 13, 2007 2:15 pm |
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Yes !!! U can. It would be the fastest approch. Please provide the answer choices & I can advise the best approcah.
Parth 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?
Can we solve this using Permutation and combination ?
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StaceyKoprince
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Post subject: Posted: Fri Dec 14, 2007 12:51 am |
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| ManhattanGMAT Staff |
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Posts: 6077
Location: San Francisco
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Please don't forget to post the full text of the question, including answer choices!
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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Guest
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Post subject: Re: Bill has a small deck of 12 playing cards Posted: Tue Dec 18, 2007 12:29 am |
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Parth 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?
Can we solve this using Permutation and combination ?
The answer choices are :
A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33
Thanks!
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StaceyKoprince
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Post subject: Posted: Tue Dec 18, 2007 2:56 am |
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| ManhattanGMAT Staff |
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Posts: 6077
Location: San Francisco
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The method shown in the official explanation is the easiest way (probability 1-X). You can use prob / comb here but you really don't want to given the other method. Did you read that explanation? Do you have any questions about it?
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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Guest
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Post subject: Posted: Tue Dec 18, 2007 1:58 pm |
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I did understand the explaination but during the test I was unable to solve this question.
:(
Is there a way to strategically guess and eliminate few answer chioces ?
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shaji
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Post subject: YES!!! There certainly is !!!! Posted: Thu Dec 20, 2007 10:42 am |
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Anonymous wrote: I did understand the explaination but during the test I was unable to solve this question.
:(
Is there a way to strategically guess and eliminate few answer chioces ?
Here it goes:
There r 3 possibilities, namely:1)All four cards are different numbers.
2) Two cards are identical numbers and the other two different
3)There are 2 identical cards.
Possibilities 1) and 2) have equal possibilities of occurence and 3) the least likely.Therfore the required probability is the sum of 2) and 3) makig it a little over 50% and the closest answer is 17/33(C).
This approach is what a 'clever manager' under conditions of stress owing to shortages of time sometimes does!!! Such approaches come very 'handy' on the GMAT.
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JAREDROC
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Post subject: Posted: Fri Dec 21, 2007 8:52 pm |
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In regards to this question. The Manhatten explanation explains to take the complement of the probability of getting no pairs on the 2, 3, and 4th cards. Taking the complement of not getting a pair on the last three cards to me seems to give the probability of getting one card that matches the first card pulled and does not include getting a match on cards not the same as the first (i.e. cards 2 & 3 matching or cards 3 & 4 matching). Can you please explain?
Thanks,
-Jared
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shaji
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Post subject: Not Quite!!! Posted: Sat Dec 22, 2007 1:28 am |
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JAREDROC wrote: In regards to this question. The Manhatten explanation explains to take the complement of the probability of getting no pairs on the 2, 3, and 4th cards. Taking the complement of not getting a pair on the last three cards to me seems to give the probability of getting one card that matches the first card pulled and does not include getting a match on cards not the same as the first (i.e. cards 2 & 3 matching or cards 3 & 4 matching). Can you please explain?
Thanks,
-Jared
Consider the three posibilities that I have cited. You will notice that 'Manhatten' is stating the same argument as I have mentioned i.e. The required Probability is 1-(Possibility1)=(Possibility 2+Possibity3)
Possibilty 1 is all the cards are different in value.
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StaceyKoprince
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Post subject: Posted: Mon Dec 24, 2007 3:52 pm |
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| ManhattanGMAT Staff |
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Posts: 6077
Location: San Francisco
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Hi, Jared
Most people get confused when thinking about probability b/c they try to do what you are doing - namely, to say that I have to account for "later" possibilities as I'm choosing cards. Don't think about it that way - it seems intuitive but will lead you astray b/c that's not how probability works. Probability is very much counter-intuitive for most people (as is combinatorics).
When you do the method we described, you are not leaving open the possibility that you are only checking the "first" card and could have a match on, say, cards 2 and 3. You are covering all possibilities.
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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virgo_rookie
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Post subject: Using an alternate method Posted: Sat Aug 16, 2008 11:50 am |
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Hi Stacey/Ron,
I was trying to use an alternate method to find the no of ways of selecting 4 cards from the deck all with diff nos, however I am getting confused behind the reasoning for the one which is the correct method. Pls help resolve the same:
Method 1: 6C4(to select the 4 different card nos between 1 & 6 to be there in the selection of 4 cards)*(2^4)(for each diff no we have 2 diff options)=15*16.
Now the no ways for total different slection of 4 cards is 12C4=33*15
The no of ways reqd= 1-(15*16)/(33*15)=1-16/33=17/33
Method2: (6C1*2)*(5C1*2)*(4C1*2)*(3C1*2)=24*15 which actually now turns out be higher than possible for no of ways.
The method2 though wrong is beyond comprehension as to why it is wrong. Kindly throw light in fallacy in reasoning/logic/concept.
Thanks.
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shaji
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Post subject: Elegant!!! Posted: Mon Aug 18, 2008 2:55 am |
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Method2 is erroneous because the order in which the cards are chosen is inconsequential as it leads to multiple of double counting. (1234=4321=2143=3214.........). In addition there is a computional error in expression in Method 2. The correct value is 360*16. To correct the double counting effect, you divide the expression by 4! which leads to 15*16 which is what you have correctly arrived at using Method1.
Method1 is elegant!!! .
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virgo_rookie
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Post subject: Thanks Posted: Sun Sep 07, 2008 1:06 am |
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Shaji, thanks for the explanation.
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chaitanya.indukuri
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Post subject: Re: Bill has a small deck of 12 playing cards Posted: Sun Sep 27, 2009 10:17 pm |
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Another method of solving this is:-
Total combinations of picking 4 cards:- 12C4
Total combinations of picking 4 cards with 2 cards with same numbers is:-
6*10C2 - 6C4
You subtract 6C4 to remove the double counting.
So, (6*10C2 - 6C4 )/ 12C4 = 17/33
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esledge
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Post subject: Re: Bill has a small deck of 12 playing cards Posted: Tue Oct 27, 2009 3:12 pm |
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| ManhattanGMAT Staff |
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Location: St. Louis, MO
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chaitanya.indukuri wrote: Another method of solving this is:-
Total combinations of picking 4 cards:- 12C4
Total combinations of picking 4 cards with 2 cards with same numbers is:-
6*10C2 - 6C4
You subtract 6C4 to remove the double counting.
So, (6*10C2 - 6C4 )/ 12C4 = 17/33 This is an interesting and clever method. I had to puzzle over it for a while to figure out where you got the 6C4, and others may do the same. 6C4 is the same as 6C2, both equal 6!/4!2! = 6*5/2 = 15. So I see 6*10C2 as (# of options for which pair is made)*(ways to select the other 2 cards from the remaining 10 cards). But you want to avoid double counting when the "other two" are a pair themselves. For example, so far, we have counted [1,1,2,2] and [2,2,1,1] as two different selections. So how many sets of 4 cards could be made of 2 matched pairs? 6 pairs are possible, and we would include 2 of them: 6C2 = 6!/2!4!
_________________
Emily Sledge
Instructor
ManhattanGMAT
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