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# M11-08

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Math Expert
Joined: 02 Sep 2009
Posts: 41872

Kudos [?]: 128641 [0], given: 12181

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16 Sep 2014, 00:44
Expert's post
8
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

64% (00:54) correct 36% (01:37) wrong based on 92 sessions

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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 positive difference between the numbers on these cards is 3 or less?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 128641 [0], given: 12181

Math Expert
Joined: 02 Sep 2009
Posts: 41872

Kudos [?]: 128641 [0], given: 12181

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16 Sep 2014, 00:44
Official Solution:

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 positive difference between the numbers on these cards is 3 or less?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

The only pair which has the positive difference more than 3 is {2,6}: $$6-2=4 \gt 3$$. Probability of this pair to be drawn is $$\frac{1}{C^2_6}$$.

So, the probability that the positive difference between the numbers on the cards is 3 or less is: $$P=1-\frac{1}{C^2_6}=\frac{14}{15}$$.

_________________

Kudos [?]: 128641 [0], given: 12181

Intern
Joined: 08 Apr 2013
Posts: 25

Kudos [?]: 13 [0], given: 6

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30 Sep 2015, 23:27
1
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Hi Bunuel,

Doesn't the use of the phrase "positive difference between the numbers on these cards" mean that we need to eliminate the possibility of picking two 5s as well. Difference between two 5s is 0, which is technically not a difference at all.

Kudos [?]: 13 [0], given: 6

Senior Manager
Joined: 31 Mar 2016
Posts: 406

Kudos [?]: 78 [0], given: 197

Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

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31 Aug 2016, 05:22
I think this is a high-quality question and I agree with explanation.

Kudos [?]: 78 [0], given: 197

Manager
Joined: 18 Jun 2016
Posts: 106

Kudos [?]: 21 [0], given: 76

Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 700 Q49 V36

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17 Dec 2016, 05:33
Bunuel wrote:
Official Solution:

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 positive difference between the numbers on these cards is 3 or less?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

The only pair which has the positive difference more than 3 is {2,6}: $$6-2=4 \gt 3$$. Probability of this pair to be drawn is $$\frac{1}{C^2_6}$$.

So, the probability that the positive difference between the numbers on the cards is 3 or less is: $$P=1-\frac{1}{C^2_6}=\frac{14}{15}$$.

Hi Bunnel,

6C2 is number of ways in which cards can be chosen, not pairs.

there are 2 ways in which this particular pair of number can be chosen (6,2 or 2,6)

So shouldn't it be 1 - 2/15 = 13/15??
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Status: Long way to go!
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17 Dec 2016, 11:20
sidoknowia wrote:
Bunuel wrote:
Official Solution:

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 positive difference between the numbers on these cards is 3 or less?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

The only pair which has the positive difference more than 3 is {2,6}: $$6-2=4 \gt 3$$. Probability of this pair to be drawn is $$\frac{1}{C^2_6}$$.

So, the probability that the positive difference between the numbers on the cards is 3 or less is: $$P=1-\frac{1}{C^2_6}=\frac{14}{15}$$.

Hi Bunnel,

6C2 is number of ways in which cards can be chosen, not pairs.
Choosing randomly 2 cards from the set means choosing pairs

there are 2 ways in which this particular pair of number can be chosen (6,2 or 2,6)
(6,2) and (2,6) are considered the same since we dont cover the order in counting

So shouldn't it be 1 - 2/15 = 13/15??

This should be 1-1/15=14/15
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Kudos [?]: 878 [0], given: 53

Intern
Joined: 29 Jul 2016
Posts: 8

Kudos [?]: [0], given: 14

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08 Jun 2017, 09:52
Bunuel wrote:
Official Solution:

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 positive difference between the numbers on these cards is 3 or less?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

The only pair which has the positive difference more than 3 is {2,6}: $$6-2=4 \gt 3$$. Probability of this pair to be drawn is $$\frac{1}{C^2_6}$$.

So, the probability that the positive difference between the numbers on the cards is 3 or less is: $$P=1-\frac{1}{C^2_6}=\frac{14}{15}$$.

Bunuel, would you please elaborate more on your answer? I initially approached this via 6!/3! to get rid of the duplicates and this gives 120, but I believe this is wrong because the problem implies that ordering does not matter because the problem states 'positive difference'. However I do not understand 6C2. What about the duplicate 5s?

A simpler example would be a list of numbers 2,3,3 and I am trying to choose two numbers. 3C2 would give me (2,3), (3,3) and (2,3) for 3 combinations, but should it not only be 2?

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22 Jul 2017, 10:29
It's not just the digits, but also the cards. 6 'distinct' cards.

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Intern
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06 Aug 2017, 18:50
Total possibilities: 15.
There is only 1 combination when positive difference is more than 3(6,2)
so the result is: 15-1=14/15

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Senior Manager
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Location: India
Concentration: Social Entrepreneurship, General Management
GMAT 1: 690 Q49 V34
GMAT 2: 720 Q49 V39
GPA: 2.8

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20 Sep 2017, 05:44
Bunuel 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 positive difference between the numbers on these cards is 3 or less?

A. $$\frac{8}{15}$$
B. $$\frac{9}{15}$$
C. $$\frac{10}{15}$$
D. $$\frac{12}{15}$$
E. $$\frac{14}{15}$$

I have a couple of doubts while solving this question .
1. if its not given the cards are distinct ,do we assume they are?
2. while eliminating the number pairs why do we not consider 6,2?

Kudos [?]: 3 [0], given: 196

Re: M11-08   [#permalink] 20 Sep 2017, 05:44
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# M11-08

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