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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=11/6*1/6=35/36 The probability of chosing 2 and 6 =1/6 each



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Re: m11,#8 [#permalink]
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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=11/6*1/6=35/36
The probability of chosing 2 and 6 =1/6 eachProbability of selecting two cards from (2,6) = 2C2 / 6C2 = 2/15 Hence, the required probability = 12/15 = 13/15.



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Re: m11,#8 [#permalink]
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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 => 12/36 = 11/18 = 17/18 ???



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Re: m11,#8 [#permalink]
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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=11/6*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 xy or yx? x = first y = last When 2 and 6 are slelcted, 62 = 4 but 26 = 4. Seems in this case order matters because if 2 comes first and 6 comes last, it should be 26 and vice versa. If I am correct, the only difference that is greater than 3 is 62 = 4. In that case, then the prob is 1  1/(6P2) = 1  1/30 = 29/30. Expect some comments.
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Re: m11,#8 [#permalink]
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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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Re: m11,#8 [#permalink]
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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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Re: m11,#8 [#permalink]
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05 Jan 2009, 09:55
Yes, answer 14/15 is correct.



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Re: m11,#8 [#permalink]
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06 Jan 2009, 22:54
14/15 is correct:)



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Re: m11,#8 [#permalink]
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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 = 11/9 = 8/9



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Re: m11,#8 [#permalink]
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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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Re: m11,#8 [#permalink]
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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 Please help why I am wrong?
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Re: m11,#8 [#permalink]
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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
Please help why I am wrong? 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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Re: m11,#8 [#permalink]
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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










