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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
30 Jul 2018, 10:58
1
Expert Reply
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Question Stats:
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?
(A) \(\frac{2}{9}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{4}{9}\)
(D) \(\frac{5}{9}\)
(E) \(\frac{2}{3}\)
(A) \(\frac{2}{9}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{4}{9}\)
(D) \(\frac{5}{9}\)
(E) \(\frac{2}{3}\)
Re: A number cube has six faces numbered 1 through 6. If the cub
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31 Jul 2018, 07:25
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 \(1 - \frac{16}{36} = \frac{20}{36} = \frac{10}{18} = \frac{5}{9}\)
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
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 \(1 - \frac{4}{9} =\frac{5}{9}\).
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 \(1 - \frac{4}{9} =\frac{5}{9}\).
