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

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Senior Manager
Joined: 22 Dec 2009
Posts: 301

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16 Feb 2010, 09:36
walker wrote:
C.

p=1-q, where q is the probability to turn over 4 cards with different values.

12P4 - the total number of combinations.

12*(11-1)*(10-2)*(9-3)=12*10*8*6 - the number of combinations with 4 cards with different values.

p=1-12*10*8*6/12P4=1-12*10*8*6*8!/12!=1-12*10*8*6/(12*11*10*9)=1-8*6/(11*9)=1-16/33=17/33

A question walker.. why do u consider 12p4 as the total combinations and not 12c4....! I guess we use P only when order matters. In this the order is not an issue as we draw 4 cards together... rite??
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16 Feb 2010, 10:35
Amardeep Sharma 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?

A 8/33
B 62/165
C 17/33
D103/165
E 25/33
Amar

p - probability of atleast one pair of cards that have the same value
q - probability of no card having same value

p = 1-q

q=(6C4*2^4)/12C4 = 16/33
p =17/33 hence C.
Manager
Joined: 15 Apr 2010
Posts: 114

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24 Oct 2010, 22:20
walker wrote:
C.

p=1-q, where q is the probability to turn over 4 cards with different values.

12P4 - the total number of combinations.

12*(11-1)*(10-2)*(9-3)=12*10*8*6 - the number of combinations with 4 cards with different values.

p=1-12*10*8*6/12P4=1-12*10*8*6*8!/12!=1-12*10*8*6/(12*11*10*9)=1-8*6/(11*9)=1-16/33=17/33

Exactly the solution I was looking for
Math Expert
Joined: 02 Sep 2009
Posts: 52296

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25 Oct 2010, 05:19
1
Amardeep Sharma 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?

A 8/33

B 62/165

C 17/33

D103/165

E 25/33

Amar

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.
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A deck of playing cards - probability  [#permalink]

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19 Mar 2011, 07:03
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

I want to know how to solve this.

OA is provided.
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Joined: 20 Dec 2010
Posts: 1809
Re: A deck of playing cards - probability  [#permalink]

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19 Mar 2011, 12:01
1
Find the opposite probability and subtract it from 1.

First card be any of the 12 cards, so P(1st card) = 12/12=1
Second card be any of the 10 cards out of the remaining 11 cards, so P(2nd card) = 10/11
Third card be any of the 8 cards out of the remaining 10 cards, so P(3rd card) = 8/10
Fourth card be any of the 6 cards out of the remaining 9 cards, so P(4th card) = 6/9

Probability of having no pair in the 4 card = 1*(10/11)*(8/10)*(6/9) = 16/33

Probability of having at least 1 pair = 1-(16/33) = 17/33

Ans: "C"

******************************************************

Let me elaborate on one of the cases;
1,1,2,2,3,3,4,4,5,5,6,6 ->12 cards
1st card-> any card can be taken: 12/12=1

Let's say we picked 4 as the first card

2nd card be anything but 4
2nd card can be: 1,1,2,2,3,3,5,5,6,6 (Total Count=10)
Total available cards = 1,1,2,2,3,3,4,5,5,6,6(Total Count=11)
P(2nd Card) = 10/11

Let's say we picked 1 as the 2nd card.

3rd card be anything but 1 or 4
3rd card can be: 2,2,3,3,5,5,6,6 (Total Count=8)
Total available cards = 1,2,2,3,3,4,5,5,6,6(Total Count=10)
P(3rd Card) = 8/10

Let's say we picked 6 as the 3rd card.

4th card be anything but 1 or 4 or 6
4th card can be: 2,2,3,3,5,5(Total Count=6)
Total available cards = 1,2,2,3,3,4,5,5,6(Total Count=9)
P(4th Card) = 6/9
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Re: A deck of playing cards - probability  [#permalink]

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20 Mar 2011, 21:07
1 - Probability(He'll not find any with same value)

1 - {12/12 * 10/11 * 8/10 * 6/9}

= 1 - {(8 * 6)/(9 * 11)}

= 1 - 16/33 = 17/33

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Joined: 27 Jan 2011
Posts: 16
Re: A deck of playing cards - probability  [#permalink]

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27 Jun 2011, 03:13
I am replying to a dormant paste, I have some issue with my approach.Someone kindly help:

Actually one way to solve this problem is calculate the complement of what is asked and subtract from 1 i.e.,{1-p}

Now i tried to solve this question by calculating the probability for atleast one pair that is (one pair + two different cards) or two paira.
A) Probability of two pairs:
first chance:12/12
Second:1/11
Third:10/10
Fourth:1/9
So (12/12)*(1/11)*(10/10)*(1/9) = 1/99

A) Probability of one pair and the other cards being different:
first chance:12/12
Second:1/11
Third:10/10
Fourth:8/9

it will come out to 8/99

Finally (A) + (B) is coming out to be 1/11

It is evident that I am missing something can someone help me out.Please!
Manager
Joined: 21 May 2011
Posts: 205
PS - 700 level - decks  [#permalink]

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04 Sep 2011, 13:14
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
Intern
Joined: 26 Sep 2010
Posts: 8
Location: United States
Schools: LBS '14
GMAT 1: 730 Q49 V40
GPA: 3.34
WE: Asset Management (Commercial Banking)
Re: PS - 700 level - decks  [#permalink]

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04 Sep 2011, 15:02
I think its simple. You just have to calculate for all the possibilities for a pair to turn up. They are as follows

First Two are a pair:
1/11

3rd card completes the pair:
10/11*2/10 = 2/11

4th card completes the pair:
10/11*8/10*3/9=8/33

Total: 1/11 + 2/11 + 8/33 = 17/33

Hope i am right.
Manager
Joined: 04 Jun 2011
Posts: 153
Re: Probability-Bill and a deck of 12 playing cards  [#permalink]

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06 Sep 2011, 07:05
farjad.... though your answer is right... im not sure of the approach... u are here considering only1 pair... while the q ays ATLEAST 1 pair... which implies... 1 and 2 can form a pair... 3 and 4 can form another pair!!

so to avoid the ambiguity.. the best approach is 1 - NO pairs

so to draw a card from the pack 1
to draw the next card which is not a pair to the first is 10/11
to draw the next u should not draw a pair of the first or the second... so u have 8/10
and to draw the card which is not pair to either of the first three is 6/9

1* 10/11*8/10*6/9 = 16/33

therefore the probability to get atleast one pair is 1-16/33 = 17/33
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Re: Bill has a small deck of 12 playing cards made up of only 2  [#permalink]

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23 Jul 2018, 09:06
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Re: Bill has a small deck of 12 playing cards made up of only 2 &nbs [#permalink] 23 Jul 2018, 09:06

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

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