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There are four cards in a box, each with a

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There are four cards in a box, each with a  [#permalink]

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New post 12 Jul 2017, 20:58
2
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (01:57) correct 34% (01:41) wrong based on 50 sessions

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There are four cards in a box, each with a different number written on it - either 1,3,4 or 5. What is the probability that the difference between the numbers on two randomly selected cards is not greater than 3?

A. 2/3
B. 3/4
C. 5/6
D. 7/8
E. 11/12

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New post 12 Jul 2017, 21:07
Not greater than 3..but can be equal to right? So there are 5 possibilities total

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New post 12 Jul 2017, 21:10
But am confused about the denominator(total)

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New post 12 Jul 2017, 21:43
pritha.k wrote:
Not greater than 3..but can be equal to right? So there are 5 possibilities total

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Yes, you cannot have 5 and 1 as the two card combination
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New post 12 Jul 2017, 21:55
pritha.k wrote:
Not greater than 3..but can be equal to right? So there are 5 possibilities total

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There are total 4C2 = 6 ways to get 2 numbers out of the given 4 numbers. The combination of 1 and 5 is the only combination that when subtracted will give result greater than 3. So, the required answer will be (6-1)/6 = 5/6.

Answer will be C.
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New post 12 Jul 2017, 22:19
Thank you for the explanation...source? OG?

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Re: There are four cards in a box, each with a   [#permalink] 12 Jul 2017, 22:19
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