Deck of cards

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Nailya: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Sat Feb 21, 2009 3:46 pm; Location: New York City; Thanked: 2 times

by Nailya » Mon Mar 30, 2009 3:40 pm

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?

[spoiler]OA 17/33[/spoiler]

Please explain how to solve this problem quickly?

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Source: Beat The GMAT — Problem Solving | Mon Mar 30, 2009 3:40 pm

mike22629: Master | Next Rank: 500 Posts; Posts: 322; Joined: Fri Mar 27, 2009 3:56 pm; Thanked: 24 times; GMAT Score:710

by mike22629 » Mon Mar 30, 2009 5:50 pm

The best way to approach this problem is to find the probability of finding no pairs and subtracting that from one.

Hence:
Probability of picking first: 1 (because it can be any card)
Probability of second card not pairing: 10/11
Probability of third card not pairing : 8/10
Probability of fourth card not pairing: 6/9

1*(10/11)(8/10)(6/9) = 16/33 (probability of no pairs)

1 - (16/33) = 17/33

Probability of at least one pair is 17/33

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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: Deck of cards

by Vemuri » Mon Mar 30, 2009 6:14 pm

Probability of selecting first card = 12/12 = 1 ( It can be any card).

Probability of selecting second card = 10/11 ( 11 cards are left for selection. We shouldn't select the matching card. )

Probability of selecting the third card = 8/10 ( 10 cards left we shouldn't select the two cards that match the already selected 2 cards)

Probability of selecting the 4th card = 6/9 ( 9 cards are left, we shouldn't select the 3 cards that match the already selected 3 cards)

Probability of selecting the cards with no pairs = 1* 10/11*8/10*6/9 = 16/33

Probability of selecting at least one pair = 1-16/33 = 17/33.

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Nailya: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Sat Feb 21, 2009 3:46 pm; Location: New York City; Thanked: 2 times

by Nailya » Wed Apr 01, 2009 9:46 am

Thank you guys!!!

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