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A 6-sided cube has faces numbered 1 through 6. If the cube i
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30 Jul 2018, 10:51
30 Jul 2018, 10:51
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Question Stats:
62% (00:58) correct
37% (01:03) wrong based on 89 sessions
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A 6-sided cube has faces numbered 1 through 6. If the cube is rolled twice, what is the probability that the sum of the two rolls is 8?
(A) \(\frac{1}{9}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{5}{36}\)
(D) \(\frac{1}{6}\)
(E) \(\frac{7}{36}\)
(A) \(\frac{1}{9}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{5}{36}\)
(D) \(\frac{1}{6}\)
(E) \(\frac{7}{36}\)
Re: A 6-sided cube has faces numbered 1 through 6. If the cube i
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07 Aug 2018, 16:50
07 Aug 2018, 16:50
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Expert Reply
Explanation
The probability of any event equals the number of ways to get the desired outcome divided by the total number of ways for the event to happen. Starting with the denominator, use the
fundamental counting principle to compute the total number of ways to roll a cube twice.
There are 6 possibilities (1, 2, 3, 4, 5, or 6) for the first roll and 6 for the second, giving a total of (6)(6) = 36 possibilities for the two rolls. For the numerator, determine the number of possible combinations that will sum to 8. For example, rolling a 2 the first time and a 6 the second time.
The full set of options is (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2). Thus, there are 5 possible combinations that sum to 8, yielding a probability of \(\frac{5}{36}\).
The probability of any event equals the number of ways to get the desired outcome divided by the total number of ways for the event to happen. Starting with the denominator, use the
fundamental counting principle to compute the total number of ways to roll a cube twice.
There are 6 possibilities (1, 2, 3, 4, 5, or 6) for the first roll and 6 for the second, giving a total of (6)(6) = 36 possibilities for the two rolls. For the numerator, determine the number of possible combinations that will sum to 8. For example, rolling a 2 the first time and a 6 the second time.
The full set of options is (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2). Thus, there are 5 possible combinations that sum to 8, yielding a probability of \(\frac{5}{36}\).
Re: A 6-sided cube has faces numbered 1 through 6. If the cube i
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21 Sep 2021, 21:24
21 Sep 2021, 21:24
Can anyone suggest an alternative to the brute force approach used here?
