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01 Dec 2023, 23:19
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D
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Question Stats:
74% (01:33) correct
26%
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The faces of a cube are labeled from 1 to 6, and when the cube is tossed, each number is equally likely to appear on the top face. If two of these cubes are tossed, what is the probability that the number that appears on the top face of either of the cubes exceeds the number on the other by exactly 2?
A. \(1\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)
photo_2023-12-02_11-42-57.jpg [ 82.94 KiB | Viewed 6000 times ]
A. \(1\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)
Attachment:
photo_2023-12-02_11-42-57.jpg [ 82.94 KiB | Viewed 6000 times ]
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02 Dec 2023, 02:08
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gmatophobia
We need the combined probability of the following pairs:
- (1, 3)
(3, 1)
(2, 4)
(4, 2)
(3, 5)
(5, 3)
(4, 6)
(6, 4)
There are 8 pairs possible from the total of 6*6=36 pairs. Therefore, the probability is 8/36 = 2/9.
Answer: D.
01 Apr 2025, 16:54
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The faces of a cube are labeled from 1 to 6, and when the cube is tossed, each number is equally likely to appear on the top face. If two of these cubes are tossed, what is the probability that the number that appears on the top face of either of the cubes exceeds the number on the other by exactly 2?
Of the 6 faces of each cube, the faces with 1, 2, 3, and 4 can be exceeded by 2 by the face on the other cube.
So, for a given cube, there's a 4/6 chance that it will show a face that can be exceeded by 2 by the face of the other cube.
Then, there's a 1/6 chance that the face on the other cube will be 2 greater than the face on the first cube.
So, if we toss one cube and then the other, the chance is 2/3 × 1/6 that we'll first get a face that can be exceeded by 2 and then get the face that exceeds the first by 2.
2/3 × 1/6 = 1/9
At the same time, there's also a 1/9 chance that the combination of faces will be achieved in the reverse order, with the face that exceeds the other by 2 showing first.
1/9 + 1/9 = 2/9
A. \(1\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)
Correct answer: D
Of the 6 faces of each cube, the faces with 1, 2, 3, and 4 can be exceeded by 2 by the face on the other cube.
So, for a given cube, there's a 4/6 chance that it will show a face that can be exceeded by 2 by the face of the other cube.
Then, there's a 1/6 chance that the face on the other cube will be 2 greater than the face on the first cube.
So, if we toss one cube and then the other, the chance is 2/3 × 1/6 that we'll first get a face that can be exceeded by 2 and then get the face that exceeds the first by 2.
2/3 × 1/6 = 1/9
At the same time, there's also a 1/9 chance that the combination of faces will be achieved in the reverse order, with the face that exceeds the other by 2 showing first.
1/9 + 1/9 = 2/9
A. \(1\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)
Correct answer: D
