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They tried to make it trickier.

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Senior Manager
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They tried to make it trickier. [#permalink]

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03 Jul 2006, 19:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

They tried to make it trickier..
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04 Jul 2006, 06:26
amansingla4, can you post an explanation as to how you got to your solution ?
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I am only one; but I am still one. I cannot do everything, but still I can do something. I will not refuse to do the something I can do.
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04 Jul 2006, 07:41
total combination: 9C3 = 84

@ least 2 Yellow:

1. comb of 2 yellow: 4C2 = 6
2. comb all yellow: 4C3 = 4
3. comb of 1 red: 2C1 = 2
4. comb of 1 brown: 3C1 = 3

6*2+6*3+4 = 34 (number of at least 2 yellow)

Exactly 1 yellow:
1. comb of 1 yellow: 4C1 = 4
2. comb of 1 red: 2C1 = 2
3. comb of 2 red: 2C2 =1
4. comb of 1 brown: 3C1 = 3
5. comb of 2 brown: 3c2=3

4*1+4*3+4*2*3=40 (number of exaclty 1 yellow)

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...

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04 Jul 2006, 10:54
acfuture wrote:

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...

B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42

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04 Jul 2006, 13:08
amartin6165 wrote:
acfuture wrote:

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...

B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42

Agree with this but 5C3 is 10 not 20.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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04 Jul 2006, 17:06
amartin6165 wrote:
B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42

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Senior Manager
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Location: Europe

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05 Jul 2006, 01:31
amartin6165 wrote:
acfuture wrote:

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...

B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42

Very unclear,
as 9C3 - 5C3 = 84 -10 = 74, and 74/84 = 37/42

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05 Jul 2006, 01:31
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They tried to make it trickier.

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