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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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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.
_________________

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???

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.

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

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17 Sep 2012, 22:33

1

This post was BOOKMARKED

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 .
_________________

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

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18 Sep 2012, 00:17

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.

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.
_________________

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

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28 Nov 2012, 06:02

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

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

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

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06 Aug 2013, 04:53

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?

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

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19 Oct 2014, 13:26

hey bunuel! i thought of this problem like this: I tried to find the probability of this guy not getting a pair, but i am not getting the required final result for this no pair case. For the first draw, number of options=12, for the second one, he can draw from 10 of the remaining(so as to not repeat the face value). for the third card, he has 8 options, and similarly 6 for the last card. So i went on to multiply these outcomes, and got 12*10*8*6. These were the number of outcomes to not get a pair. Now i divided it by 12C4, to get the probability.(to get the final answer i will surely subtract it from 1) But for the answer, the denominator is taken only as 12*11*10*9, i don't get it as to why it is not 12C4 , what am i missing? thanks in advance

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

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01 May 2015, 05:52

IanStewart wrote:

shrouded1 wrote:

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.

Hi

Can you please correct me where am I going wrong in the below approach

1-P(no pair) P(no pair) = (10x8x6)/12C4

Assuming that the first card can be any among them, let it be 1,2,3,4,5 or 6. Its a fixed value. So the number of ways of selecting it is one. After that the number of ways of selecting the second card is 10 and so on.

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

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01 May 2015, 05:54

cumulonimbus wrote:

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

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

Answer: C.

Hope it helps.

Hi Bunnel

Can you please correct me where am I going wrong in the below approach

1-P(no pair) P(no pair) = (10x8x6)/12C4

Assuming that the first card can be any among them, let it be 1,2,3,4,5 or 6. Its a fixed value. So the number of ways of selecting it is one. After that the number of ways of selecting the second card is 10 and so on.

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.

Hi

Can you please correct me where am I going wrong in the below approach

1-P(no pair) P(no pair) = (10x8x6)/12C4

Assuming that the first card can be any among them, let it be 1,2,3,4,5 or 6. Its a fixed value. So the number of ways of selecting it is one. After that the number of ways of selecting the second card is 10 and so on.

Please correct me where i am going wrong.

Thanks!!

You are mixing concepts. You have 12 distinct cards. There are two different methods of solving this:

Method 1: The "number of ways" of selecting the first card is 12, not 1. The number of ways of selecting the second card is 10. The number of ways of selecting the third card is 8 and the number of ways of selecting the fourth card is 6. Now you have the 4 cards such that there are no pairs. What are the total number of ways of selecting 4 cards? They are 12*11*10*9 Required Probability = 1 - (12*10*8*6)/(12*11*10*9) = 17/33

Note that you have arranged cards in first, second, third and fourth places. It doesn't matter as long as you arrange them in the denominator as well which you did by using 12*11*10*9 and not 12C4.

Method 2: 1 is the probability of selecting the first card such that there are no pairs. It can be any card so probability of picking it is 1. In this case the second card probability is 10/11 and so on... Required Probability = 1 - 1*(10/11)*(8/10)*(6/9) = 17/33
_________________

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

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04 May 2015, 09:14

Hi

Can you please correct me where am I going wrong in the below approach

1-P(no pair) P(no pair) = (10x8x6)/12C4

Assuming that the first card can be any among them, let it be 1,2,3,4,5 or 6. Its a fixed value. So the number of ways of selecting it is one. After that the number of ways of selecting the second card is 10 and so on.

Please correct me where i am going wrong.

Thanks!![/quote]

You are mixing concepts. You have 12 distinct cards. There are two different methods of solving this:

Method 1: The "number of ways" of selecting the first card is 12, not 1. The number of ways of selecting the second card is 10. The number of ways of selecting the third card is 8 and the number of ways of selecting the fourth card is 6. Now you have the 4 cards such that there are no pairs. What are the total number of ways of selecting 4 cards? They are 12*11*10*9 Required Probability = 1 - (12*10*8*6)/(12*11*10*9) = 17/33

Note that you have arranged cards in first, second, third and fourth places. It doesn't matter as long as you arrange them in the denominator as well which you did by using 12*11*10*9 and not 12C4.

Method 2: 1 is the probability of selecting the first card such that there are no pairs. It can be any card so probability of picking it is 1. In this case the second card probability is 10/11 and so on... Required Probability = 1 - 1*(10/11)*(8/10)*(6/9) = 17/33[/quote]

Hi Karishma

Thanks for the explanation.

If i get you right, by mixing of concepts you meant that when I am considering the first card can be any of the card so it's number of ways should be one is wrong. That methodology can be applied only when I am calculating the probability, not the number of ways to select it ?

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

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13 Jun 2015, 20:44

Hi Guys ,can someone please tell me why this method would be incorrect? To select 4 cards from 2 different suits with each having different values ,I could go the following ways a)Select 2 each from Suit A and Suit B -> 6c2*4c2(To get rest 2 cards from the 4 different values of the other suit) b)3 from A and 1 from B-> 6c3*3c1 c) 1 from A and 3 from B->6c1*5c3 d)All four from A -> 6c4 e)All four from B->6c4 when I add them all up I get the correct answer but I'm not sure if this is the correct approach..Please guide me!! Thanks

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

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13 Jun 2016, 23:16

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

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

Answer: C.

Hope it helps.

Although I do understand your approach but not sure what's wrong with mine.!

I did this as follows -> P = P(one pair) + P(2 pairs)

6C1*10C1*8C1/12C4 + 6C2/12C4

6C1 - no. of ways of selecting 1 pair out of 6 pairs 10C1 - no. of ways of selecting 1 card out of the remaining 10 8C1 - no. of ways of selecting 1 card out of the remaining 8 cards ( excluding the 9th card as that would form another pair which is considered in the second case) 12C4 - total no. of ways of selecting 4 cards out of 12 cards 6C2 - no. of ways of selecting 2 pairs out of 6 pairs

I am not getting the right answer - not sure which point I am missing. Can you please explain? Thanks.!

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

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

Answer: C.

Hope it helps.

Although I do understand your approach but not sure what's wrong with mine.!

I did this as follows -> P = P(one pair) + P(2 pairs)

6C1*10C1*8C1/12C4 + 6C2/12C4

6C1 - no. of ways of selecting 1 pair out of 6 pairs 10C1 - no. of ways of selecting 1 card out of the remaining 10 8C1 - no. of ways of selecting 1 card out of the remaining 8 cards ( excluding the 9th card as that would form another pair which is considered in the second case) 12C4 - total no. of ways of selecting 4 cards out of 12 cards 6C2 - no. of ways of selecting 2 pairs out of 6 pairs

I am not getting the right answer - not sure which point I am missing. Can you please explain? Thanks.!

Here is your problem:

6C1*10C1*8C1

6C1 - no. of ways of selecting 1 pair out of 6 pairs ------ fine 10C1 - no. of ways of selecting 1 card out of the remaining 10 -------- say you picked the 2 of spades 8C1 - no. of ways of selecting 1 card out of the remaining 8 cards ---------say you picked the 3 of hearts

Now, in 10C1, you will pick the 3 of hearts in another case and in 8C1, you will pick the 2 of spades. Hence, there is double counting.

Divide 10C1 * 8C1 by 2 to get

(6C1*10C1*8C1)/(2*12C4) + 6C2/12C4

You should get the correct answer.
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Re: Bill has a small deck of 12 playing cards [#permalink]

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17 Sep 2016, 03:21

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?

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

Another Way

Total No of ways to Pick 4 cards out of 12 = \(C^4_{12}\)

Prob of picking atleast 1 pair = 1 - Prob of No pair.

Total No. of ways to pick no Pairs = \(\frac{(12 * 10 * 8 * 6)}{4!}\) in above expression, 12 => There are 12 ways to pick 1st card 10 => There are 10 ways to pick 2nd card because 1 has already been picked and we can't pick the same number again. 8 & 6 => Same reason as that of 10. 4! => (12 * 10 * 8 * 6) gives us the Number of ways to Pick and Arrange 4 cards. So, we divide by 4! to remove the Arrangement as their sequence doesn't matter to us

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