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

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Manager
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Bill has a small deck of 12 playing cards made up of only 2 [#permalink] New post 20 Oct 2007, 12:16
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?


8/33

62/165

17/33

103/165

25/33
VP
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 [#permalink] New post 20 Oct 2007, 12:31
I'm getting 17/33

1*10/11*8/10*6/9 = 48/99

1-48/99 = 1-16/33 = 17/33

1 ---> choosing any card

10/11 ---> not choosing his pair out of 11 cards.

8/10 ---> now you have two cards. not choosing their pairs.

6/9 ---> you have choosen three cards ! not choosing their pairs.

:)
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 [#permalink] New post 20 Oct 2007, 14:40
nice explanation killersquirrel.

im convinced of 17/33 as well
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 [#permalink] New post 12 Dec 2007, 07:31
KillerSquirrel wrote:
I'm getting 17/33

1*10/11*8/10*6/9 = 48/99

1-48/99 = 1-16/33 = 17/33

1 ---> choosing any card

10/11 ---> not choosing his pair out of 11 cards.

8/10 ---> now you have two cards. not choosing their pairs.

6/9 ---> you have choosen three cards ! not choosing their pairs.

:)


How would your answer change it it said at least TWO pairs will the same value?
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 [#permalink] New post 15 Dec 2007, 08:09
bmwhype2 wrote:
KillerSquirrel wrote:
I'm getting 17/33

1*10/11*8/10*6/9 = 48/99

1-48/99 = 1-16/33 = 17/33

1 ---> choosing any card

10/11 ---> not choosing his pair out of 11 cards.

8/10 ---> now you have two cards. not choosing their pairs.

6/9 ---> you have choosen three cards ! not choosing their pairs.

:)


How would your answer change it it said at least TWO pairs will the same value?


Since you have four cards the most you can have is two pairs (can never get three pairs in four cards !!) so you really ask for "what is the probability for him to choose two pairs" (i.e 2,2 and 4,4).

(1*1/11*1*1/9) + (1*10/11*2/10*1/9) + (1*1/11*1*1/9) = 1/99 + 2/99 + 1/99 = 4/99

:)
Manager
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 [#permalink] New post 15 Dec 2007, 14:02
KillerSquirrel wrote:
bmwhype2 wrote:
KillerSquirrel wrote:
I'm getting 17/33

1*10/11*8/10*6/9 = 48/99

1-48/99 = 1-16/33 = 17/33

1 ---> choosing any card

10/11 ---> not choosing his pair out of 11 cards.

8/10 ---> now you have two cards. not choosing their pairs.

6/9 ---> you have choosen three cards ! not choosing their pairs.

:)


How would your answer change it it said at least TWO pairs will the same value?


Since you have four cards the most you can have is two pairs (can never get three pairs in four cards !!) so you really ask for "what is the probability for him to choose two pairs" (i.e 2,2 and 4,4).

(1*1/11*1*1/9) + (1*10/11*2/10*1/9) + (1*1/11*1*1/9) = 1/99 + 2/99 + 1/99 = 4/99

:)


Why the 2nd ((1*1/11*1*1/9))?
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 [#permalink] New post 16 Dec 2007, 11:52
Skewed wrote:
KillerSquirrel wrote:
bmwhype2 wrote:
KillerSquirrel wrote:
I'm getting 17/33

1*10/11*8/10*6/9 = 48/99

1-48/99 = 1-16/33 = 17/33

1 ---> choosing any card

10/11 ---> not choosing his pair out of 11 cards.

8/10 ---> now you have two cards. not choosing their pairs.

6/9 ---> you have choosen three cards ! not choosing their pairs.

:)


How would your answer change it it said at least TWO pairs will the same value?


Since you have four cards the most you can have is two pairs (can never get three pairs in four cards !!) so you really ask for "what is the probability for him to choose two pairs" (i.e 2,2 and 4,4).

(1*1/11*1*1/9) + (1*10/11*2/10*1/9) + (1*1/11*1*1/9) = 1/99 + 2/99 + 1/99 = 4/99

:)


Why the 2nd ((1*1/11*1*1/9))?


Assume

2 - 2 - 4 - 4

or

2 - 4 - 4 - 2

or

4 - 4 - 2 - 2

:)
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 [#permalink] New post 16 Dec 2007, 14:55
But there are 6 variations of 2 2's and 2 4's.
2244
2424
2442
4422
4242
4224

4!/(2!2!)=6

Why are only 3 listed?
  [#permalink] 16 Dec 2007, 14:55
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