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M11-08

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Bunuel
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M11-08 [#permalink] New post   16 Sep 2014, 00:44
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55% (hard)

Question Stats:

62% (01:49) correct 38% (02:24) wrong based on 118 sessions
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There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

A. \(\frac{8}{15}\)
B. \(\frac{9}{15}\)
C. \(\frac{10}{15}\)
D. \(\frac{12}{15}\)
E. \(\frac{14}{15}\)
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E
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axezcole
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Re: M11-08 [#permalink] New post   13 Nov 2019, 23:33
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Total conditions-6p2
Condition that satisfy difference greater than 3= (2,6) (6,2)

Prob=1-2/30

Prob=14/15
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Re M11-08 [#permalink] New post   16 Sep 2014, 00:44
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Official Solution:

There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

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


The only pair with a difference greater than 3 is {2, 6}: \((6 - 2 = 4) > 3\). The probability of drawing this pair is \(\frac{1}{C^2_6}\).

Thus, the probability that the absolute 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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Re: M11-08 [#permalink] 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] New post   18 Oct 2019, 07:33
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The probability approach is as follows.

6,2 can be arranged in 2 ways

\(1-2*\frac{1}{6}*\frac{1}{5}\)
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Re: M11-08 [#permalink] New post   06 May 2023, 03:43
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M11-08 [#permalink]
06 May 2023, 03:43
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Bunuel
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