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Does this Question look like GMAT Question?
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lunarpower
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i think this is one of our questions.
in any case, this is certainly not an abnormally long problem.
if you solve this problem by direct computation, then it can take a long time. however, there's a shortcut.
the key is to notice that "at least one card" is a COMPLEX EVENT, but the OPPOSITE event ("NO matching cards") is a SIMPLE event. therefore, if we can calculate the probability of the simple event, we can just subtract from 1 to find the probability of the complex event.
the probability of the OPPOSITE event:
draw the first card --> 12/12 (any first card will do)
draw the second card --> 10/11 (any card except for the one matching the first card drawn)
draw the third card --> 8/10 (any card except the two that match the ones already drawn)
draw the fourth card --> 6/9 (any card except the three that match the ones already drawn)
therefore, the probability of this opposite event is (12/12)(10/11)(8/10)(6/9), which reduces to (1/11)(8/1)(2/3), or 16/33.
therefore, the probability of the ORIGINAL event is 1 - 16/33, or 17/33.
--
there are plenty of official problems that are longer than this one. for instance, see this problem:
https://www.manhattangmat.com/forums/gprep-2-t1737.html
in any case, this is certainly not an abnormally long problem.
if you solve this problem by direct computation, then it can take a long time. however, there's a shortcut.
the key is to notice that "at least one card" is a COMPLEX EVENT, but the OPPOSITE event ("NO matching cards") is a SIMPLE event. therefore, if we can calculate the probability of the simple event, we can just subtract from 1 to find the probability of the complex event.
the probability of the OPPOSITE event:
draw the first card --> 12/12 (any first card will do)
draw the second card --> 10/11 (any card except for the one matching the first card drawn)
draw the third card --> 8/10 (any card except the two that match the ones already drawn)
draw the fourth card --> 6/9 (any card except the three that match the ones already drawn)
therefore, the probability of this opposite event is (12/12)(10/11)(8/10)(6/9), which reduces to (1/11)(8/1)(2/3), or 16/33.
therefore, the probability of the ORIGINAL event is 1 - 16/33, or 17/33.
--
there are plenty of official problems that are longer than this one. for instance, see this problem:
https://www.manhattangmat.com/forums/gprep-2-t1737.html
Ron has been teaching various standardized tests for 20 years.
--
Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
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lunarpower
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@ Musiq
incidentally, you will never see an authentic gmat problem that REQUIRES the use of combination or permutation formulas (although they certainly may help in some cases).
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i will grant you that some of the words used in the problem statement (such as "he likes to play a game") are a bit nonstandard, but this type of problem is certainly within the bailiwick of the gmat.
in any case, the hallmark of the gmat is that problems that appear complicated in one way may actually turn out to be simple when approached from a different angle.
so, if you think that a problem is unnecessarily complicated, try a different approach. don't just criticize the problem.
hopefully you have been teaching this to your 9+ years of past students.
check out my post above. this problem can be solved via simple multiplication of consecutive probabilities; it doesn't really test combinations/permutations at all, much less to an extent that would be prohibitive for the gmat.Musiq wrote:I have been teaching the GMAT for 9+ years now.
In my opinion, this is a poor representation of what the GMAT is. The GMAT tests concepts of Perms, Combs etc...but not to the extent this questions is asking and definitely not in the uninspired , non-standardized language like the question you quoted uses.
incidentally, you will never see an authentic gmat problem that REQUIRES the use of combination or permutation formulas (although they certainly may help in some cases).
--
i will grant you that some of the words used in the problem statement (such as "he likes to play a game") are a bit nonstandard, but this type of problem is certainly within the bailiwick of the gmat.
in any case, the hallmark of the gmat is that problems that appear complicated in one way may actually turn out to be simple when approached from a different angle.
so, if you think that a problem is unnecessarily complicated, try a different approach. don't just criticize the problem.
hopefully you have been teaching this to your 9+ years of past students.
Ron has been teaching various standardized tests for 20 years.
--
Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
--
Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
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goelmohit2002
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Thanks Ron !lunarpower wrote:i think this is one of our questions.
in any case, this is certainly not an abnormally long problem.
if you solve this problem by direct computation, then it can take a long time. however, there's a shortcut.
the key is to notice that "at least one card" is a COMPLEX EVENT, but the OPPOSITE event ("NO matching cards") is a SIMPLE event. therefore, if we can calculate the probability of the simple event, we can just subtract from 1 to find the probability of the complex event.
the probability of the OPPOSITE event:
draw the first card --> 12/12 (any first card will do)
draw the second card --> 10/11 (any card except for the one matching the first card drawn)
draw the third card --> 8/10 (any card except the two that match the ones already drawn)
draw the fourth card --> 6/9 (any card except the three that match the ones already drawn)
therefore, the probability of this opposite event is (12/12)(10/11)(8/10)(6/9), which reduces to (1/11)(8/1)(2/3), or 16/33.
therefore, the probability of the ORIGINAL event is 1 - 16/33, or 17/33.
--
there are plenty of official problems that are longer than this one. for instance, see this problem:
https://www.manhattangmat.com/forums/gprep-2-t1737.html
What is your opinion about using the approach that I followed....copy pasting the same again...do you think this approach is good enough to solve this problem
Divide the 12 cards in 6 groups...
(1,1), (2,2), (3,3),(4,4), (5,5), (6,6)
Selecting four different groups = 6C4
Out of these groups we have to select one card each so that none of the number match = 2C1 * 2C1 * 2C1 * 2C1
Thus total ways so that none of the cards match = 6C4 * 2C1 * 2C1 * 2C1 * 2C1
Total ways of selection of 4 cards = 12C4
Thus Probabilty when none match = say X = (6C4 * 2C1 * 2C1 * 2C1 * 2C1) / (12C4)
Required Probabilty when atleast one match = 1-X = 17/33 = C.
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lunarpower
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well, sure, of course this is a legitimate solution.goelmohit2002 wrote: What is your opinion about using the approach that I followed....copy pasting the same again...do you think this approach is good enough to solve this problem
Divide the 12 cards in 6 groups...
(1,1), (2,2), (3,3),(4,4), (5,5), (6,6)
Selecting four different groups = 6C4
Out of these groups we have to select one card each so that none of the number match = 2C1 * 2C1 * 2C1 * 2C1
Thus total ways so that none of the cards match = 6C4 * 2C1 * 2C1 * 2C1 * 2C1
Total ways of selection of 4 cards = 12C4
Thus Probabilty when none match = say X = (6C4 * 2C1 * 2C1 * 2C1 * 2C1) / (12C4)
Required Probabilty when atleast one match = 1-X = 17/33 = C.
it's not a solution that i would teach to random students, because it requires a VERY impressive grasp of the use of combination formulas. (the ability to put together a solution such as that one in <2 min means that you know the combination formulas, and their use, so well that it's probably like doing times tables for you.)
but sure, it works.
Ron has been teaching various standardized tests for 20 years.
--
Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
--
Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
