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Bill has a small deck of 12 playing cards [#permalink]

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18 Jun 2010, 22:47

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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, 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?

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 - \(\frac{6}{9}\);

Bunuel, very nice explanation. I believe, the second method can be derived logically and the first one with some knowledge of equations in permutations and combination. Thanks once again.
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bunuel, what's your chain of thought when you solve combination/permutation and probability questions? it seems to me that your logic is pretty clear when you solve this kind of problems. please share your secret/experience if you don't mind. thanks!

Re: Experts attention - Tricky Probability question [#permalink]

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13 Sep 2010, 06:30

gmatrant 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?

a.8/33 b.62/165 c.17/33 d.103/165 e.25/33

My approach-

We need to calculate 1-p(x) where x is the probability of not drawing identical numbered cards. 1.From 12 cards we can pick any one number as the first card - No of ways =12 2.We now have 10 ways of picking the next card that does not have identical number = 10 ways 3. Picking up the 3rd card in 8 ways since 2 cards are already chosen and we cannot choose the same numbered card. 4. Similarly as in 3 we can now have 6 ways to choose the fourth card

Total ways of not picking even one identical number pair card = 12*10*8*6 Total ways of picking up 4 cards out of 12 = 12 * 11 * 10 *9 Probability of x is (12*10*8*6) / (12*11*10*9) = 16/33 Hence p(1-x) = 17/33.

I have a question here The total way of picking up 4 cards our of 12 is 12C4 and it will give me a different number than I calculated above. Can you please tell me where I am going wrong here, though I get the answer.

In method 1, the order of picking matters In method 2, when you use the C(n,r) formulae, the order of picking does not matter As a result method 2 will always yield a lower number of ways than method 1 Eg. In method 1, you will count the set Suit1No1, Suit1No2,Suit2No3,Suit2No4, 4! times, one for each ordering between the 4 cards that exists, whereas in method 2, you will count it only once.

It is essentially the difference between P(n,r) and C(n,r)

This subtlety is present in a lot of probability questions and must be clarified in the Q. The above question is in MGMAT (I got it on my CAT), but it fails to tell you if the 4 cards that are picked are picked simultaneously (i.e. order does not matter) or one by one. In my opinion such questions are ambiguous and the answer is dependent on the assumption you make. Unfortunately there are more Qs like this one on the MGMAT CATs, just something to be weary of.
_________________

This subtlety is present in a lot of probability questions and must be clarified in the Q. The above question is in MGMAT (I got it on my CAT), but it fails to tell you if the 4 cards that are picked are picked simultaneously (i.e. order does not matter) or one by one. In my opinion such questions are ambiguous and the answer is dependent on the assumption you make. Unfortunately there are more Qs like this one on the MGMAT CATs, just something to be weary of.

Whether you pick the four cards simultaneously, or pick them one at a time (without replacement) doesn't actually matter if you are finding a probability; the two situations are mathematically identical. You can see this intuitively by thinking of taking hold of four cards in the deck first. If you take them all out at the same time, or if there is a nanosecond between your removing each, why would the probability that you get a pair be affected? It won't be, so the 'ambiguity' you suggest is present in such questions is no ambiguity at all.

You can see that either perspective will give you the same answer, though it's easier to illustrate with a simpler example. Say you have 3 red marbles and 4 blue marbles in a bag, and you pick two (either simultaneously, or without replacement - it's the same thing), and you want to find the probability of picking two red marbles. If we look at the problem as though we are picking marbles one at a time, we have 3*2 ways of picking two reds, and 7*6 ways of picking two marbles, so the probability would be 3*2/7*6 = 1/7. If we look at the problem as though we're picking two marbles simultaneously, we have 3C2 ways of picking two red marbles and 7C2 ways of picking two marbles, so the probability would be 3C2/7C2 = 1/7. So when you stick your two hands in the bag and grab two marbles, it doesn't matter if you lift your two hands out at the same time, or take them out one at a time; the probability is the same.

Note though that you need to be consistent in the calculation - if you assume order matters when you calculate the numerator, you must also assume order matters when you calculate the denominator.
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Re: Experts attention - Tricky Probability question [#permalink]

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13 Sep 2010, 07:19

You are right, what I should have said is that the ambiguity occurs if you are solving the combinatorial problem, where you are counting outcomes. In calculating probability it is irrelevant as the ordering terms will cancel themselves out.

In the question mentioned, the problem is if you calculate the number of relevant ways using one assumption and the total number of ways using another assumption
_________________

Bunuel - I understand the 6c4 x 2^4 approach.. but why is this wrong: 6c2 - 2 cards from 1st suite 4c2 - 2 cards from the 4 remaining (the other 2 have the same #s as the ones selected from 1st suite).. so 6c2 x 4c2?

There is 2 major problems with this :

(a) By choosing 2 cards from one suit and 2 from the other, you are forcing that the choice has exactly 2 cards from each suit, which is not true

(b) Even if you had to calculate no of ways to pick 4 cards such that no pairs and 2 cards from each suite, this is wrong, because it needs to be multiplied by 2 [2 ways to choose which is the "first suite"]
_________________

yes can see it now.. is there any other method to get the answer other than 6c4 2^4?

Using combinatorics, you can make up any number of ways to solve a problem. The essence here is counting a certain subset of scenarios, how you choose to count that is entirely up to you.

As done above you can use a "Universe - Everything Else" or you can count directly in a number of ways like "Ways to pick exactly 1 pair + Ways to pick exactly 2 pairs", "Ways to pick pairs <= 3 + ways to pick pairs with number b/w 4 &6"

The point being how to count, can be answered in an arbitrary number of ways, but the end result is the same : ALL ROADS LEAD TO ROME .. you just need to pick the shortest and easiest one. In scenarios like this one "universe-everything" gives the most elegant and straight forward solution
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Help solving a probability problem using combinatorics [#permalink]

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26 Nov 2011, 20:41

Guys,

I got this question in one of the MGMAT practice tests. The solution provided used 1-x technique. For probability questions, I want to practice at least 2 ways of solving the problem. And I believe, this problem can be solved by Combinatorics (12C4 = 495 for the denominator and anagram for the numerator). I just cant get the numerator right. Can anyone please tell me how to solve this question using Combinatorics?

---------------------------------------------------------------------------------------------------- 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
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You can see from the number of "Kudos Given" on my profile that I am generous while giving Kudos. Is it too much to ask for a similar favor???

It's a lot more work to do the problem directly than it is to find the probability of no pairs and subtract from 1, however.
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Re: Help solving a probability problem using combinatorics [#permalink]

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26 Nov 2011, 21:32

Thanks Ian. I agree its not the best way solve this particular problem. But just trying get my head around both the techniques of solving these questions. Who knows, I might run into a problem slightly different for which the 2nd technique is more appropriate

Thanks a lot for the response.
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Re: Bill has a small deck of 12 playing [#permalink]

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08 May 2012, 06:39

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total no of ways of choosing any 4 cards out of 12 = 12 C 4 =45*11

the no favourable outcomes, i.e atleast 1 pair , = no of pairs i.e 6 *{ ( no of ways of choosing the remaining 2 cards, i.e. 10 C 2 - the no of repetitions (i.e.5)} + the no ways of getting 2 pairs .i.e. 15

actually i don't care what has Bill and what size it is - but it took me more than two min to find it out compared to trillion +-1 doesnt make any difference ) but ,anyway, thank you for an advice - a quite good one!

Re: Bill has a small deck of 12 playing cards [#permalink]

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09 Aug 2012, 05:26

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

Hello, I'm strongly interested how Bunuel got the numerator in his first solution. I tried the anagramm method to determine the desired outcomes:

Cards: | 1 | 2 | 3 | 4 | 5 | 6 | 1 | 2 | 3 | 4 | 5 | 6 | Anagramm: | I | I | I | I | N | N | N | N | N | N | N | N | (I= In, N= Not in)

Thus, I get \(\frac{(6!*6!)}{(4!*8!)}\) I'm a bit confused, because we calculate the total outcomes by counting the posibilities of choosing 4 cards out of 12. The desired outcomes must be calculated by counting the posibilities by choosing 4 diferent values out of 6 and multiplying the distribution of these posibilities with the posibilities to arrange the 2 decks.

What I mean is \(\frac{no. of values}{no. of cards}\) could not be reduced?! As I said, I am confused!

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.
_________________

- Stne

Last edited by stne on 17 Sep 2012, 22:28, edited 2 times in total.

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 made up of only 2
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17 Sep 2012, 10:32

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