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A number cube has six faces numbered 1 through 6. If the cub
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30 Jul 2018, 10:58
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53% (01:25) correct
46% (01:41) wrong based on 66 sessions
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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?
Re: A number cube has six faces numbered 1 through 6. If the cub
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31 Jul 2018, 07:25
2
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 \(1 - \frac{16}{36} = \frac{20}{36} = \frac{10}{18} = \frac{5}{9}\)
Re: A number cube has six faces numbered 1 through 6. If the cub
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09 Aug 2018, 14:32
1
Expert Reply
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 \(1 - \frac{4}{9} =\frac{5}{9}\).
Re: A number cube has six faces numbered 1 through 6. If the cub
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15 Oct 2022, 22:06
Given that A number cube has six faces numbered 1 through 6 and We need to find 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?
As we are rolling the cube twice => Number of cases = \(6^2\) = 36
P(at least one of the rolls will result in a number greater than 4) = 1 - P(None of the two rolls will result in a number greater than 4)
P(None of the two rolls will result in a number greater than 4) = P(Both the rolls will have a number from 1 to 4)
Following are the possible cases: (1,1), (1,2), (1,3), (1,4) (2,1), (2,2), (2,3), (2,4) (3,1), (3,2), (3,3), (3,4) (4,1), (4,2), (4,3), (4,4)
=> 16 cases
=> P(None of the two rolls will result in a number greater than 4) = \(\frac{16}{36}\) = \(\frac{4}{9}\)
=> P(at least one of the rolls will result in a number greater than 4) = 1 - P(None of the two rolls will result in a number greater than 4) = 1 - \(\frac{4}{9}\) = \(\frac{9 - 4}{9}\) = \(\frac{5}{9}\)
So, Answer will be D Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
gmatclubot
Re: A number cube has six faces numbered 1 through 6. If the cub [#permalink]