A Deck of Cards

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A Deck of Cards

by csandeepreddy » Sun Oct 19, 2008 3:12 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?

A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33

Can someone really solve this problem under two minutes?
Source: — Problem Solving |

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by Ian Stewart » Sun Oct 19, 2008 6:47 pm
Yes, it is possible to solve quite quickly in fact, as long as you flip the problem around: if you figure out the probability you do *not* get a pair, then subtract from 1, you find the probability of getting at least one pair. See Sudhir's solution here (and I offered an alternative direct solution in the following discussion):

www.beatthegmat.com/combination-and-per ... t8924.html
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by cramya » Sun Oct 19, 2008 8:18 pm
Ian,
Saw sudhir's solution.
In some problems I gte stumped on when to multiply veruss when to add.

For example in Sudhir's solution

thus the total probability of not picking any matching card is

10/11*8/10*6*9 =16/33


I can understand this is an AND situatiion so multiply. Is that the rule of thumb? Please advice.

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by vittalgmat » Mon Oct 20, 2008 5:47 am
Ian Stewart wrote:Yes, it is possible to solve quite quickly in fact, as long as you flip the problem around: if you figure out the probability you do *not* get a pair, then subtract from 1, you find the probability of getting at least one pair. See Sudhir's solution here (and I offered an alternative direct solution in the following discussion):

www.beatthegmat.com/combination-and-per ... t8924.html
Ian,
On a totally tangential note, can one expect to see Q which uses deck of cards as an example ?? . I am kinda curious. I agree that cards provide lot of interesting problems in probability, permutation, combn. However, I think, for a card question, the test maker has to assume that the test taker knows about them for him to refrain from mentioning them, lest the problem description would become very wordy. ( eg 4 suites of 13 cards each, blah blah). On the other hand, explaining dice in 1 sentence is easy.

Ian, stuart, stacey, ron.. your thoughts please.

thanks
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by sudhir3127 » Mon Oct 20, 2008 6:55 am
cramya wrote:Ian,
Saw sudhir's solution.
In some problems I gte stumped on when to multiply veruss when to add.

For example in Sudhir's solution

thus the total probability of not picking any matching card is

10/11*8/10*6*9 =16/33


I can understand this is an AND situatiion so multiply. Is that the rule of thumb? Please advice.
ramya .. u can use the following link ..

https://mathforum.org/library/drmath/view/56503.html

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by Ian Stewart » Mon Oct 20, 2008 7:24 am
vittalgmat wrote:
Ian Stewart wrote:Yes, it is possible to solve quite quickly in fact, as long as you flip the problem around: if you figure out the probability you do *not* get a pair, then subtract from 1, you find the probability of getting at least one pair. See Sudhir's solution here (and I offered an alternative direct solution in the following discussion):

www.beatthegmat.com/combination-and-per ... t8924.html
Ian,
On a totally tangential note, can one expect to see Q which uses deck of cards as an example ?? . I am kinda curious. I agree that cards provide lot of interesting problems in probability, permutation, combn. However, I think, for a card question, the test maker has to assume that the test taker knows about them for him to refrain from mentioning them, lest the problem description would become very wordy. ( eg 4 suites of 13 cards each, blah blah). On the other hand, explaining dice in 1 sentence is easy.

Ian, stuart, stacey, ron.. your thoughts please.

thanks
-V
If you did see a question about a standard deck of cards on the GMAT, they would need to explain what's in a deck of a cards- 4 suits, thirteen types of card, etc. Otherwise people who don't play cards would be at a disadvantage. I'm pretty sure I've seen a cards question, but it had nothing to do with playing cards- the cards were numbered from 1 to 10 or something similar.

If you take an undergrad combinatorics (counting/probability) course, you'll see dozens of questions about cards; one of the standard exercises is working out the probability of various poker hands. Working out, say, the probability of being dealt a full house with five cards is too difficult for the GMAT, but could be good practice for the highest level counting problems on the test - at least, if you can do that problem, you can probably to most GMAT counting problems.

cramya- yes, if you want to determine the probability that two independent events both occur, you multiply their individual probabilities. If you still have questions after reading the link above, post here and I'll do my best to answer.
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by cramya » Mon Oct 20, 2008 9:35 am
Thanks Sudhir and Ian! I appreciate the help!

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by vittalgmat » Mon Oct 20, 2008 12:26 pm
Thanks Ian and ramya