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M11-08

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M11-08 [#permalink]

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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

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Re M11-08 [#permalink]

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New post 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: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).


Answer: E
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M11-08 [#permalink]

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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.

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Re M11-08 [#permalink]

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

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Re: M11-08 [#permalink]

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New post 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: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).


Answer: E


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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Re: M11-08 [#permalink]

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New post 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: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).


Answer: E


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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Re: M11-08 [#permalink]

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New post 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: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).


Answer: E


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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Re: M11-08 [#permalink]

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

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Re: M11-08 [#permalink]

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New post 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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Re: M11-08 [#permalink]

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New post 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?

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