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You are given a deck of standard playing cards with 13 of

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You are given a deck of standard playing cards with 13 of [#permalink] New post 16 Jul 2003, 01:01
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D
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You are given a deck of standard playing cards with 13 of the 52 cards designated as a тАЬheart.тАЭ You plan to shuffle the deck thoroughly then deal 10 cards off the top of the deck. What is the probability that the 10th card dealt is a heart?
(A) 1/4
(B) 1/5
(C) 5/26
(D) 12/42
(E) 13/42
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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

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 [#permalink] New post 16 Jul 2003, 03:49
i didnt quite get the problem.

Probabilty of getting a heart is 1/4 always in a pack of 52 cards.

The other option which may make sense is C.

Shall try to post a soultion again... :(
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 [#permalink] New post 16 Jul 2003, 03:53
evensflow wrote:
i didnt quite get the problem.

Probabilty of getting a heart is 1/4 always in a pack of 52 cards.

The other option which may make sense is C.

Shall try to post a soultion again... :(


The problem is exactly as written (I try to be careful when writting these things to make sure there is no ambiguity as to assumptions).

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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

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 [#permalink] New post 16 Jul 2003, 04:35
We need:

(1) first nine nonheart cards and then a heart one

OR

(2) first nine any cards and then a heart one
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 [#permalink] New post 16 Jul 2003, 13:11
Okay, here is a hint. Try solving analytically for 2 cards, and maybe 3 cards and see if you can find a generalization.
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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

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Heart Problem [#permalink] New post 16 Jul 2003, 13:18
Isn't it more complex than that. Isn't it the probability of

1 heart in first nine (probably of 10th card being a heart is 12/43) or
2 hearts in first nine (probably of 10th card being a heart is 11/43) or
3 hearts in first nine (probably of 10th card being a heart is 10/43) or
4 hearts ....or 9 hearts in first nine.

I can't seem to figure out the answer though. Because I get 85 in the numerator....

Another hint?
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 [#permalink] New post 16 Jul 2003, 13:33
Ladies and Gentlemen!

Take a card! Any card! Is it a heart?

:wink:
_________________

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

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Joined: 07 Jul 2003
Posts: 771
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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Kudos [?]: 32 [0], given: 0

 [#permalink] New post 16 Jul 2003, 21:20
The correct answer is (A). Although this may be counter-intuitive at first, the probability of any card in the deck being a heart before any cards are seen is 1/4. We can also solve this analytically for any card by building a probability тАЬtreeтАЭ and summing the probability of all of its тАЬbranches.тАЭ

For example, letтАЩs find the probability that the 2nd card is a heart. There are two mutually exclusive ways that can happen: (1) both the first and second cards are hearts; and (2) only the second card is a heart.
Using the multiplication rule, the probability that the first card is a heart AND the second card is a heart is equal to the probability of picking a heart on the first card or 13/52 (number of hearts in a full deck divided by number of cards in the deck) times the probability of subsequently picking a heart on the 2nd card or 12/51 (number of hearts remaining in the deck divided by number of cards remaining in the deck) which equals 12/204.

Similarly, the probability that the first card is a non-heart AND that the second card is a heart is equal to the probability that the first card in NOT a heart or 39/52 (number of non-hearts in a new deck divided by number of cards in the deck) times the probability of subsequently picking a heart on the 2nd card or 13/51 (number of hearts in the deck divided by the number of cards remaining in the deck) or 39/204.

Since these two events are mutually exclusive, we can add them together to get the total probability of getting a heart as the second card: i.e., 12/204 + 39/204 = 51/204 = 1/4. We can do a similar analysis for any card in the deck, and, although the probability tree gets more complicated as the card number gets higher, the total probability of the nth card being a heart will always end up simplifying to 1/4.
_________________

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

  [#permalink] 16 Jul 2003, 21:20
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