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# Can anybody give a good solution to this one? An eight-card

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Founder
Joined: 04 Dec 2002
Posts: 15585

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Location: United States (WA)
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Can anybody give a good solution to this one? An eight-card [#permalink]

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28 Jul 2003, 19:42
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can anybody give a good solution to this one?

An eight-card deck consists of 4 pairs of each suit. 2 cards are taken at random without replacement. What is the probability to have two cards of the same suit?
тАв 1/2
тАв 1/3
тАв 1/4
тАв 1/7
тАв 1/8

Last edited by bb on 28 Jul 2003, 19:53, edited 1 time in total.

Kudos [?]: 28510 [0], given: 5114

GMAT Instructor
Joined: 07 Jul 2003
Posts: 769

Kudos [?]: 235 [0], given: 0

Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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28 Jul 2003, 21:05
bb wrote:
Can anybody give a good solution to this one?

An eight-card deck consists of 4 pairs of each suit. 2 cards are taken at random without replacement. What is the probability to have two cards of the same suit?
тАв 1/2
тАв 1/3
тАв 1/4
тАв 1/7
тАв 1/8

The absolute quickest way to solve this is to say:

Whatever card I pick as the first card, there is only one card of the seven remaining to match its suit. Hence, the answer is 1/7
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Kudos [?]: 235 [0], given: 0

Founder
Joined: 04 Dec 2002
Posts: 15585

Kudos [?]: 28510 [0], given: 5114

Location: United States (WA)
GMAT 1: 750 Q49 V42

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28 Jul 2003, 22:37
AkamaiBrah wrote:
bb wrote:
Can anybody give a good solution to this one?

An eight-card deck consists of 4 pairs of each suit. 2 cards are taken at random without replacement. What is the probability to have two cards of the same suit?
тАв 1/2
тАв 1/3
тАв 1/4
тАв 1/7
тАв 1/8

The absolute quickest way to solve this is to say:

Whatever card I pick as the first card, there is only one card of the seven remaining to match its suit. Hence, the answer is 1/7

Very good reasoining!!!

You are the man!

Kudos [?]: 28510 [0], given: 5114

Manager
Joined: 22 Jul 2003
Posts: 61

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Location: CA

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29 Jul 2003, 07:58
Another way would be:

Favorable outcome = 4 (because there are only 4 cases where both the cards will be of same suite)

Total outcome = 8C2 = 28

P = 4/28 = 1/7
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P K Das

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Intern
Joined: 27 Apr 2003
Posts: 36

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29 Jul 2003, 13:02
great soluton akamai ..

you are a star

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29 Jul 2003, 13:02
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