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# A 6-sided cube has faces numbered 1 through 6. If the cube i

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Joined: 07 Jun 2014
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GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1622 [0], given: 385

A 6-sided cube has faces numbered 1 through 6. If the cube i [#permalink]  30 Jul 2018, 10:51
Expert's post
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Question Stats:

43% (00:25) correct 56% (00:30) wrong based on 16 sessions
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) $$\frac{7}{36}$$
[Reveal] Spoiler: OA

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Sandy
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GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4720
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 91

Kudos [?]: 1622 [0], given: 385

Re: A 6-sided cube has faces numbered 1 through 6. If the cube i [#permalink]  07 Aug 2018, 16:50
Expert's post
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 $$\frac{5}{36}$$.
_________________

Sandy
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Re: A 6-sided cube has faces numbered 1 through 6. If the cube i   [#permalink] 07 Aug 2018, 16:50
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