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A number cube has six faces numbered 1 through 6. If the cub

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GMAT Club Legend
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A number cube has six faces numbered 1 through 6. If the cub [#permalink] New post 30 Jul 2018, 10:58
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Question Stats:

45% (00:49) correct 54% (01:27) wrong based on 11 sessions
A number cube has six faces numbered 1 through 6. If the cube is rolled twice, what is the probability that at least one of the rolls will result in a number greater than 4?

(A) \(\frac{2}{9}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{4}{9}\)
(D) \(\frac{5}{9}\)
(E) \(\frac{2}{3}\)
[Reveal] Spoiler: OA

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Re: A number cube has six faces numbered 1 through 6. If the cub [#permalink] New post 31 Jul 2018, 07:25
The probability of both roll being 4 or less than 4 is \(\frac{4}{6} * \frac{4}{6} = \frac{16}{36}\)
Therefore the probability of a roll being more than 4 is \(1 - \frac{16}{36} = \frac{20}{36} = \frac{10}{18} = \frac{5}{9}\)
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Re: A number cube has six faces numbered 1 through 6. If the cub [#permalink] New post 09 Aug 2018, 14:32
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Explanation

Because this problem is asking for an “at least” solution, use the 1 – x shortcut.

The probability that at least one roll results in a number greater than 4 is equal to 1 minus the probability that both of the rolls result in numbers 4 or lower. For one roll, there are 6 possible outcomes (1 through 6) and 4 ways in which the outcome can be 4 or lower, so the probability is \(\frac{4}{6}=\frac{2}{3}\).

Thus, the probability that both rolls result in numbers 4 or lower is \(\frac{2}{3}\frac{2}{3}=\frac{4}{9}\).

This is the result that you do not want; subtract this from 1 to get the probability that you do want. The probability that at least one of the rolls results in a number greater than 4 is \(1 - \frac{4}{9} =\frac{5}{9}\).
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Re: A number cube has six faces numbered 1 through 6. If the cub   [#permalink] 09 Aug 2018, 14:32
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