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# m11,#8

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11 Nov 2008, 23:35
There are six cards bearing numbers 2,4,5,5,5,6. If two cards are randomly selected from the lot, what is the probability that the difference between the numbers on these cards is 3 or less?

I solved this question in this way but got answer wrong What is wrong with my method?

Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - probability to select 2 and 6=1-1/6*1/6=35/36

The probability of chosing 2 and 6 =1/6 each
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11 Nov 2008, 23:54
ritula wrote:
There are six cards bearing numbers 2,4,5,5,5,6. If two cards are randomly selected from the lot, what is the probability that the difference between the numbers on these cards is 3 or less?

I solved this question in this way but got answer wrong What is wrong with my method?

Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - probability to select 2 and 6=1-1/6*1/6=35/36

The probability of chosing 2 and 6 =1/6 each

Probability of selecting two cards from (2,6) = 2C2 / 6C2 = 2/15
Hence, the required probability = 1-2/15 = 13/15.
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31 Dec 2008, 21:02
If i were to use brute force method:-

as there is 6 cards, so there will be 6 x 6 = 36 different results

As there is only 2 possibility of getting a difference of more than 3: (6,2) & (2,6)

Why is it the probability not => 1-2/36 = 1-1/18 = 17/18 ???

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01 Jan 2009, 00:23
ritula wrote:
There are six cards bearing numbers 2,4,5,5,5,6. If two cards are randomly selected from the lot, what is the probability that the difference between the numbers on these cards is 3 or less?

I solved this question in this way but got answer wrong What is wrong with my method?

Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - probability to select 2 and 6=1-1/6*1/6=35/36

The probability of chosing 2 and 6 =1/6 each

Seems the question is little ambigious: how the difference is calculated? Is it x-y or y-x?
x = first
y = last

When 2 and 6 are slelcted, 6-2 = 4 but 2-6 = -4. Seems in this case order matters because if 2 comes first and 6 comes last, it should be 2-6 and vice versa.

If I am correct, the only difference that is greater than 3 is 6-2 = 4.
In that case, then the prob is 1 - 1/(6P2) = 1 - 1/30 = 29/30.

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04 Jan 2009, 08:12
here is my question:

P(3 or less) =# of ways to select a 2,4 or 5/ all possible outcomes , right ?

all possible outcomes are given by 6C2. Number of ways to choose a 2,4,or 5 is given by 5C2, isnt it ? Where am i going wrong in my logic ?
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04 Jan 2009, 17:25
I differ with most of the answer and even few explanation...below is my answer and explanation....correct me, if anything

Approach 1:
Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - 1/6C2 = 1 - 1/ 15 = 14/15
(as there is only one such combination that has diff more than 3 i.e. (2,4))

Approach 2:
How many such combinations are there which has diff of <= 3...there are 14...you need to manually find out...(2,4), (2,5), (2,5), (2,5), (4,5), etc..
So, Probability that the difference is 3 or less = 14/ 6C2 = 14/15

Obviously Approach 1 is much faster...choosing right approach with some quick intelligence saves lot of time in exam....
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05 Jan 2009, 09:55
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06 Jan 2009, 22:54
14/15 is correct:)
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08 Jan 2009, 21:40
gmattarget700 wrote:
I differ with most of the answer and even few explanation...below is my answer and explanation....correct me, if anything

Approach 1:
Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - 1/6C2 = 1 - 1/ 15 = 14/15
(as there is only one such combination that has diff more than 3 i.e. (2,4))

Approach 2:
How many such combinations are there which has diff of <= 3...there are 14...you need to manually find out...(2,4), (2,5), (2,5), (2,5), (4,5), etc..
So, Probability that the difference is 3 or less = 14/ 6C2 = 14/15

Obviously Approach 1 is much faster...choosing right approach with some quick intelligence saves lot of time in exam....

I have question regarding approach 1.

If (2,4) = (4,2) and is considered only 1 combination...then the total combination is:

(2,2) (2,4) (2,5) (2,6) (4,4) (4,5) (4,6) (5,5) (5,6) = 9 combination

As (2,4) only appear once, then by right the answer = 1-1/9 = 8/9
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05 Jul 2010, 16:59
I also got 14/15 by just looking at it .
But after looking at gmattiger's approach I am secondguessing myself.
I believe gmattiger's answer is right.
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24 Sep 2010, 01:47
Guys, could you please look at my file

According to my calculations I receive 17/18 =(34/36) which is different from 14/15

Attachments

a.png [ 3.19 KiB | Viewed 1495 times ]

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11 Oct 2010, 23:29
Pkit wrote:
Guys, could you please look at my file

According to my calculations I receive 17/18 =(34/36) which is different from 14/15

pkit here the total number of possibilities are not 36 but 30.( you have considered 6 sets which are (2,2),(4,4),(5,5),(6,6).. and these sets shouldn't be considered)

so a total of 30 sets of values are present. There is one and only one possibility that the difference is greater than 3 that is the set (6,2) we cannot consider (2,6) bcause the value becomes a negative and less than 3.

so IMO the ans is 29/30..

do correct if i am wrong !!
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15 Apr 2012, 23:35
GMAT TIGER has a valid point. Does anyone else agree with him? The OA is 14/15.
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16 Apr 2012, 02:30
Question will be revised for the next edition of GMAT Club Tests.
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Re: m11,#8   [#permalink] 16 Apr 2012, 02:30
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# m11,#8

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