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# Bill has a small deck of 12 playing cards

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Re: Bill has a small deck of 12 playing cards made up of only 2 [#permalink]  17 Sep 2012, 04:28
The way that is given here I understand.
pairs-twins-and-couples-103472.html#p805725

But I am trying to do in another way , wonder what I am doing wrong.
at least one pair means either one pair or two pairs
so lets suppose we have A1 A2 A3 A4 A5 A6 AND B1 B2 B3 B4 B5 B6

I understand the 1 - p( opposite event )

but just for understanding sake , if I tried the direct way , how could it be done.

P( One pair) + p( two pairs )

one pair

6C1*5C2(4!/2!)= 720

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical

for two pairs

6C2 * 4!/2!2! = 90

6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .

hence total ways for at least one pair -> 6C1*5C2(4!/2!) + 6C2 * 4!/2!2! = 810 ( what is wrong here)
however this will not the required answer, can anybody please correct my logic and show me how to approach this direct way ?
total ways 4 cards can be selected 12C4 = 495 hence I am getting a probability more than one which is not possible

Would highly appreciate any help.
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Last edited by stne on 17 Sep 2012, 21:28, edited 2 times in total.
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Re: Bill has a small deck of 12 playing cards made up of only 2 [#permalink]  17 Sep 2012, 09:32
1
KUDOS
stne wrote:
The way that is given here I understand.
pairs-twins-and-couples-103472.html#p805725

But I am trying to do in another way , wonder what I am doing wrong.
at least one pair means either one pair or two pairs
so lets suppose we have A1 A2 A3 A4 A5 A6 AND B1 B2 B3 B4 B5 B6

I understand the 1 - p( opposite event )

but just for understanding sake , if I tried the direct way , how could it be done.

P( One pair) + p( two pairs )

one pair

6C1*10C2(4!/2!)= 720

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical

for two pairs

6C2 * 4!/2!2! = 90

6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .

hence total ways for at least one pair -> 6C1*10C2(4!/2!) + 6C2 * 4!/2!2! = 810 ( what is wrong here)
however this will not the required answer, can anybody please correct my logic and show me how to approach this direct way ?
total ways 4 cards can be selected 12C4 = 495 hence I am getting a probability more than one which is not possible

Would highly appreciate any help.

One pair - should be 6C1*5C2*2*2 = 6*10*4 = 240. Choose one pair out of 6, then two single cards from two different pairs. You have two suits, so for the same number two possibilities.
Two pairs - just 6C2=15.
Total - 255
Probability 255/495 = 17/33.

10C2 also includes remaining pairs of cards with the same number, so choosing 2 out of 10 does not guarantee two non-identical numbers.
Another problem is that 495 represents the number of choices for 4 cards, regardless to the order in which they were drawn. Then you should consider the other choices accordingly, and disregard the order in which they were chosen.
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Re: Bill has a small deck of 12 playing cards [#permalink]  17 Sep 2012, 21:33
Just edited my original post
Meant to take 5C2 not 10C2,

so my doubt

one pair

6C1*5C2(4!/2!)= 720

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical

for two pairs

6C2 * 4!/2!2! = 90

6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .
_________________

- Stne

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Re: Bill has a small deck of 12 playing cards [#permalink]  17 Sep 2012, 22:25
1
KUDOS
Expert's post
stne wrote:
Just edited my original post
Meant to take 5C2 not 10C2,

so my doubt

one pair

6C1*5C2(4!/2!)= 720

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical

for two pairs

6C2 * 4!/2!2! = 90

6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .

Responding to a pm:

There is a difference between this question and the other one you mentioned.

In that question, you were using one digit twice which made them identical. Here the two cards that form the pair are not identical. They are of different suits. Also, here, you don't need to arrange them. You can assume to just take a selection while calculating the cases in the numerator and the denominator. The probability will not get affected. In the other question, you needed to find the number of arrangements to make the passwords/numbers and hence you needed to arrange the digits.

Hence, in this step, 6C1*5C2(4!/2!), it should be 6C1*5C2*2*2 instead.

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different values from the 5 remaining to form the singles *This is correct *
2*2 = For each of the two values, you can select a card in 2 ways (since you have 2 suits)

Similarly, 6C2 * 4!/2!2! should be 6C2 only to select 2 pairs out of 6.

Probability = 255/495 = 17/33

Note: You can arrange the cards too and will still get the same probability. Just ensure that you arrange in numerator as well as denominator.

Only one pair = 6C1*5C2*2*2 * 4! (you multiply by 4! because all the cards are distinct)
Both pairs = 6C2 * 4! (you multiply by 4! because all the cards are distinct)

Select 4 cards out of 12 = 12C4 * 4! (you multiply by 4! because all the cards are distinct)

Probability = 255*4!/495*4! = 17/33
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Veritas Prep Reviews Director Joined: 22 Mar 2011 Posts: 610 WE: Science (Education) Followers: 63 Kudos [?]: 409 [1] , given: 43 Re: Bill has a small deck of 12 playing cards [#permalink] 17 Sep 2012, 22:43 1 This post received KUDOS stne wrote: Just edited my original post Meant to take 5C2 not 10C2, so my doubt one pair 6C1*5C2(4!/2!)= 720 6C1 = ways to select the one card from 6 which will form the pair 5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct * 4!/2! = permutations of 4 letters where two are identical for two pairs 6C2 * 4!/2!2! = 90 6C2 = ways to select the two cards which will form the pair . 4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind . one pair 6C1*5C2(4!/2!)= 720 NO Choose one pair out of 6 - 6C1 - and you don't care for the order in which you choose the two cards Choose two pairs out of the remaining 5 pairs - 5C2 - and then, from each pair, you have 2 possibilities to choose one of them, therefore 5C2*2*2 Here you stop! Don't care about any order. Period. for two pairs 6C2 * 4!/2!2! = 90 NO You choose two pairs - 6C2 - and you stop here. From each pair you take both cards, and don't care about any order. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Manager Joined: 27 May 2012 Posts: 208 Followers: 0 Kudos [?]: 46 [0], given: 73 Re: Bill has a small deck of 12 playing cards [#permalink] 17 Sep 2012, 23:17 VeritasPrepKarishma wrote: Responding to a pm: There is a difference between this question and the other one you mentioned. In that question, you were using one digit twice which made them identical. Here the two cards that form the pair are not identical. They are of different suits. Also, here, you don't need to arrange them. You can assume to just take a selection while calculating the cases in the numerator and the denominator. The probability will not get affected. In the other question, you needed to find the number of arrangements to make the passwords/numbers and hence you needed to arrange the digits. Hence, in this step, 6C1*5C2(4!/2!), it should be 6C1*5C2*2*2 instead. 6C1 = ways to select the one card from 6 which will form the pair 5C2 = select two different values from the 5 remaining to form the singles *This is correct * 2*2 = For each of the two values, you can select a card in 2 ways (since you have 2 suits) Similarly, 6C2 * 4!/2!2! should be 6C2 only to select 2 pairs out of 6. Probability = 255/495 = 17/33 Awesome , this was going to be my next question, I was going to ask that the singles can be selected first and then the pair or we could have different arrangements as the question does not say that the pair should be together ( adjacent)so we could have B3A1A5A2 where A1A2 is the pair of a single suit and B3A5 are the single cards. Below explanation already clarifies that question. Thank you for hitting the bulls eye. This was exactly what was bothering me. VeritasPrepKarishma wrote: Note: You can arrange the cards too and will still get the same probability. Just ensure that you arrange in numerator as well as denominator. Only one pair = 6C1*5C2*2*2 * 4! (you multiply by 4! because all the cards are distinct) Both pairs = 6C2 * 4! (you multiply by 4! because all the cards are distinct) Select 4 cards out of 12 = 12C4 * 4! (you multiply by 4! because all the cards are distinct) Probability = 255*4!/495*4! = 17/33 This was enlightening . EvaJager wrote: one pair 6C1*5C2(4!/2!)= 720 NO Choose one pair out of 6 - 6C1 - and you don't care for the order in which you choose the two cards Choose two pairs out of the remaining 5 pairs - 5C2 - and then, from each pair, you have 2 possibilities to choose one of them, therefore 5C2*2*2 Here you stop! Don't care about any order. Period. for two pairs 6C2 * 4!/2!2! = 90 NO You choose two pairs - 6C2 - and you stop here. From each pair you take both cards, and don't care about any order. Thank you Eva .. But the question I was going to ask next to you was that, " the pair can be selected first or the singles can be selected first or the pair can be in between the 2 singles etc " , the question does not say that the pair has to be together and singles together or that that the pair has to be selected first and then the singles , so this thought was confusing me. I was basically getting confused with this sum, digit-codes-combination-103081.html#p802805 Karishma has very clearly answered that very doubt.Just wanted to know " why I was doing , what I was doing , ". Its now clear, thanks to both of you , hope I am in a better position to understand such questions now. Karishma and Eva thank you so much for this. _________________ - Stne Intern Joined: 19 Feb 2012 Posts: 17 Followers: 0 Kudos [?]: 1 [0], given: 1 Re: Bill has a small deck of 12 playing cards [#permalink] 28 Nov 2012, 05:02 Hi Bunuel No doubt that your method is right, but this is my thought process an Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A the number of patterns that A,B,C,A hand appears = 4C2 Therefore, number of ways A is chosen= 12 when the first A is chosen, it locks in the value for the second A number of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list) number of ways C is chosen = 9 P(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33 Adhil Manager Joined: 06 Jun 2010 Posts: 161 Followers: 2 Kudos [?]: 16 [0], given: 151 Re: Bill has a small deck of 12 playing cards [#permalink] 05 Mar 2013, 11:34 Hi Bunuel, My doubt here is that: 6C4 # of ways to choose 4 different cards out of 6 different values; 2 ^4 -as each of 4 cards chosen can be of 2 different suits. Im not clear why we did 6C4 even though we selected out of 12 cards and also the purpose of doing 2^4. Thanks, Shreeraj Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 857 Kudos [?]: 3611 [1] , given: 144 Re: Probability [#permalink] 29 Apr 2013, 02:41 1 This post received KUDOS Expert's post Bunuel wrote: maheshsrini 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, for each value from 1 to 6, there are two cards in the deck with that 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? 8/33 62/165 17/33 103/165 25/33 Let's calculate the opposite probability ans subtract this value from 1. Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}. Responding to a pm: How do we obtain 6C4*2^4? Think of what you have: 6 cards numbered 1 to 6 of 2 different suits. Say you have 1 to 6 of clubs and 1 to 6 of diamonds. A total of 12 cards. You want to select 4 cards such that there is no pair i.e. no two cards have the same number. This means all 4 cards will have different numbers, say, you get a 1, 2, 4 and 6. The 1 could be of clubs or diamonds. The 2 could be of clubs and diamonds and so on for all 4 cards. This means you must select 4 numbers out of the 6 numbers in 6C4 ways. Then for each number, you must select a suit out of the given two suits. That is how you get 6C4 * 2*2*2*2 This is the total number of ways in which you will have no pair i.e. two cards of same number. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Probability [#permalink]  26 May 2013, 06:09
Bunuel wrote:
maheshsrini 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, for each value from 1 to 6, there are two cards in the deck with that 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?
8/33
62/165
17/33
103/165
25/33

Let's calculate the opposite probability ans subtract this value from 1.

Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}.

C^4_6 - # of ways to choose 4 different cards out of 6 different values;
2^4 - as each of 4 cards chosen can be of 2 different suits;
C^4_{12} - total # of ways to choose 4 cards out of 12.

So P=1-\frac{16}{33}=\frac{17}{33}.

Or another way:

We can choose any card for the first one - \frac{12}{12};
Next card can be any card but 1 of the value we'v already chosen - \frac{10}{11} (if we've picked 3, then there are one more 3 left and we can choose any but this one card out of 11 cards left);
Next card can be any card but 2 of the values we'v already chosen - \frac{8}{10} (if we've picked 3 and 5, then there are one 3 and one 5 left and we can choose any but these 2 cards out of 10 cards left);
Last card can be any card but 3 of the value we'v already chosen - \frac{6}{9};

P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}.

So P=1-\frac{16}{33}=\frac{17}{33} - the same answer as above.

Hope it helps.

Hi Bunnel,

I used combinatorics, is this way correct.

Prob (at least 1 pair) = fav/tot
fav = 12c1*1c1*10c10*9c8 + 12c1*1c1*10c10*1c1
tot = 12c4

Prob = (12*10*9/8+12*10) / (12*11*10*9/4*3*2)
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Re: Bill has a small deck of 12 playing cards [#permalink]  06 Aug 2013, 03:53
I solved this question as follows and I know I am wrong. The problem is: I don't know why am I wrong?

First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4

Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4

Could somebody point out the error in this solution and why?
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A game of cards [#permalink]  16 Dec 2013, 02:01
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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?

1.8/33
2.62/165
3.17/33
4.103/165
5.25/33
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Re: A game of cards [#permalink]  16 Dec 2013, 02:04
Expert's post
vishalrastogi wrote:
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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?

1.8/33
2.62/165
3.17/33
4.103/165
5.25/33

Merging similar topics. Please refer to the solutions on page 1.

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Re: A game of cards   [#permalink] 16 Dec 2013, 02:04
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