Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2912:30 AM EDT
-01:30 AM EDT
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Apr 3001:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
01 Dec 2023, 23:19
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (01:33) correct
26%
(01:42)
wrong
based on 528
sessions
History
Date
Time
Result
Not Attempted Yet
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 5998 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 5998 times ]
Using official mocks already?
Add GMAT Club Tests for harder practice, analytics, and explanations without overlapping Official Guide or official mock questions.
Try it now →
02 Dec 2023, 02:08
Kudos
Bookmarks
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
Kudos
Bookmarks
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
