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Bill has a small deck of 12 playing cards made up of only 2 suits of 6

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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 14 Jun 2015, 22:21
1
davesinger786 wrote:
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


Yes, using this method, you will get the number of ways in which you will have no pairs. When you subtract those from the total number of ways, you will get the number of ways in which you will have at least one pair. This will give you the required probability. Nothing wrong here. This method will be a bit time consuming though, hence I would suggest you to check out Bunuel's approach given above too.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 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 - \(\frac{6}{9}\);

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

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

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.!
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 14 Jun 2016, 21:53
MeghaP 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 - \(\frac{6}{9}\);

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

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

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 made up of only 2 suits of 6  [#permalink]

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New post 06 May 2017, 17:15
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


We can use the equation:

1 = P(at least one pair of cards in 4) + P(no pairs of cards in 4)

Let’s determine the probability of not finding a pair of cards when 4 are flipped.

P(first flip) = 1

P(second flip is not a pair with the first card) = 10/11

P(third card is not a pair with the first or second card) = 8/10

P(fourth card is not a pair with the first, second, or third card) = 6/9

Thus, the probability of not selecting a pair is:

10/11 x 8/10 x 6/9

10/11 x 4/5 x 2/3

2/11 x 4 x 2/3

16/33

Since 1 = P(at least one pair of cards in 4) + P(no pairs of cards in 4),

P(at least one pair of cards in 4) = 1 - P(no pairs of cards in 4) = 1 - 16/33 = 17/33.

Answer: C
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 01 Jan 2018, 03:44
Heres a simple and cool and innovative method. :-D

Bill chooses 2 cards with at least 1 pair. Therefore the problem can be solved in the following manner:-

a) total possibilities = 12x11x10x9 cards
b) If he choose at least 1 pair, then there are 2
Case1: Where he chooses exactly 1 pair = 12x10x8x3 (Using the Bunuel method, I simply replaced 6 by 3 because at least 1 of the 3 numbers is
Case2: Where he chooses exactly 2 pairs = 12x10x2x1 (Uptil 10, then he has only 2 cards)

Probability = (12x10x8x3 + 12x10x2x1)/12x11x10x9 = 26/99.

But the answer is not correct. I fail to understand why? I have 25 short on the numerator.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 22 Feb 2018, 13:25
Can someone please explain the 2^4 part in original solution.

If Say, I have to choose 2 numbers from 5, say the chosen numbers are 1 and 2 , then the way I can include is (1,2) (2,1) I can not include (1,1) and (2,2) how can there be four ways to get 1 and 2 from 2 suits(someone gave this ans as a solution of selecting one pair by 6c1*5c2x2^2) where I am assuming that method of getting 2^2 is same as getting 2^4.

Help will be appreciated... Thanks

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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 23 Feb 2018, 01:18
Mudit27021988 wrote:
Can someone please explain the 2^4 part in original solution.

If Say, I have to choose 2 numbers from 5, say the chosen numbers are 1 and 2 , then the way I can include is (1,2) (2,1) I can not include (1,1) and (2,2) how can there be four ways to get 1 and 2 from 2 suits(someone gave this ans as a solution of selecting one pair by 6c1*5c2x2^2) where I am assuming that method of getting 2^2 is same as getting 2^4.

Help will be appreciated... Thanks

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You are looking for the ways in which you have no pair. So you will select 4 numbers out of 6 (from 1 to 6). This is done in 6C4 ways. Say we get 1, 2, 4, 5 in one case.
But we have 2 suits say spades and hearts. The 1 can be of either spades or hearts (it will lead to 2 different cases). Similarly, 2 can be either spades or hearts and so on...
So for each number, you have 2 options of the suit. Hence you multiply by 2 * 2 * 2 * 2 = 2^4.
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 23 Feb 2018, 01:42
Hi VeritasPrepKarishma
Thanks for the reply. Like in your example you mentioned 1,2,4,5.so as u rightly mentioned that each could come from different suits. So aren't we counting (1,1) here which is not required.

We have already eliminates the probability of getting (3,3) (6,6) likewise should we subtract some number from 2^4.

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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 24 Feb 2018, 05:23
Mudit27021988 wrote:
Hi VeritasPrepKarishma
Thanks for the reply. Like in your example you mentioned 1,2,4,5.so as u rightly mentioned that each could come from different suits. So aren't we counting (1,1) here which is not required.

We have already eliminates the probability of getting (3,3) (6,6) likewise should we subtract some number from 2^4.

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No, we cannot have (1, 1). Note that we have 4 distinct numbers already - 1, 2, 4 and 5. Each will appear once only.
But we can have two cases -
1 of spades, 2 of spades, 4 of hearts and 5 of spades
1 of hearts, 2 of spades, 4 of hearts and 5 of spades

So by multiplying by 2, we are accounting for these 2 cases. Same goes for each number. That is why we have 2*2*2*2
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6  [#permalink]

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New post 16 Sep 2018, 01:13
I think there is a different way to solve this Question.
12C4 = 495

at least 1 pair = 2c2 *6 * 10c2 -5 = 6*(45-5) = 240 ( -5 for to avoid any remaining pair)
At least 2 pair: 6c2 = 15

Total possibilities = 15 + 240=255

255/495 = 17/33
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Re: Bill has a small deck of 12 playing cards made up of only 2 suits of 6   [#permalink] 16 Sep 2018, 01:13

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