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

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Bill has a small deck of 12 playing cards made up of only 2 [#permalink] New post 05 Oct 2007, 22:39
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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, 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

D. 103/165

E. 25/33

Last edited by singh_amit19 on 05 Oct 2007, 22:44, edited 1 time in total.
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 [#permalink] New post 05 Oct 2007, 22:42
My approach:

1 - probability of non-similar cards being picked up
= 1- 2 X (6C4/12C4)
= 31 / 33
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 [#permalink] New post 05 Oct 2007, 23:03
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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Re: PS - MGMAT CAT 1 [#permalink] New post 05 Oct 2007, 23:05
singh_amit19 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

D. 103/165

E. 25/33


I get C.
lets first find the probability of no pairs then we will subtract it from 1.

1st card . He can pick any so probability 1

second card. he has 11 and he can pick any 10 of them apart from the one which is similar to the previous one already picked so probability is 10/11

third card. He can pick 8 of the 10 remaining. leaving the 2 already picked.so probability 8/10

fourth card. similar as above it will be 6/9

so total probability of no pairs is

10/11 * 8/10 * 6/9 = 16/33. there to get atleast one matching pair, probability is 1- 16/33 = 17/33
Re: PS - MGMAT CAT 1   [#permalink] 05 Oct 2007, 23:05
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