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globaldesi
Joined: 28 Jul 2016
Last visit: 23 Feb 2026
Posts: 1,140
Given Kudos: 67
Location: India
Concentration: Finance, Human Resources
Schools: ISB '18 (D)
GPA: 3.97
WE:Project Management (Finance: Investment Banking)
Products:
27 Apr 2021, 05:23
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oranges +marbles = 7 (total)
after an orange is taken total sample left =6
lets assume an orange si taken
4/6
so total prob
5/7*4/6
20/42
after an orange is taken total sample left =6
lets assume an orange si taken
4/6
so total prob
5/7*4/6
20/42
27 Apr 2021, 05:58
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Bookmarks
jinroayeee
a) Sum will be ≤5 in the following cases
(1,3) (1,4) (2,3) = 3 cases
Total cases of two numbers = \(^3C_1 * ^3C_1 = 9\)
Probability \(= 1-\frac{3}{9} = \frac{2}{3}\)
b) Product will be ≤5 in the following cases
(1,3) (1,4) (1,5) = 3 cases
Total cases of two numbers = \(^3C_1 * ^3C_1 = 9\)
Probability \(= 1-\frac{3}{9} = \frac{2}{3}\)
1) Please post one question per post
2) Please post questions with options
27 Apr 2021, 09:29
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Hello, jinroayeee. I like the process that GMATinsight has outlined above. If a more rigorous approach eludes you, you can also resort to testing the possibilities and using logic and pattern recognition to solve each problem, since the sets are small enough.
a) 1 + 3 = 4; 1 + 4 = 5; 1 + 5 = 6 → 1 desirable sum
2 + 3 = 5; 2 + 4 = 6; 2 + 5 = 7 → 2 desirable sums (it should be obvious that the next three will work)
3 + 3 = 6; 3 + 4 = 7; 3 + 5 = 8 → 3 desirable sums
6 desirable sums out of 9 possibilities is
\(\frac{6}{9} = \frac{2}{3}\)
b) 1 * 3 = 3; 1 * 4 = 4; 1 * 5 = 5 → 0 desirable products
2 * 3 = 6; 2 * 4 = 8; 2 * 5 = 10 → 3 desirable products (it should be obvious that the next three will work)
3 * 3 = 9; 3 * 4 = 12; 3 * 5 = 15 → 3 desirable products
6 desirable products out of 9 possibilities is
\(\frac{6}{9} = \frac{2}{3}\)
The worst thing you can do when you see a question is seize up and become afraid to make a move. Even a step in the right direction can get things going sometimes, and an elementary way of solving a question is just as valid as any other way.
Good luck with your studies.
- Andrew
a) 1 + 3 = 4; 1 + 4 = 5; 1 + 5 = 6 → 1 desirable sum
2 + 3 = 5; 2 + 4 = 6; 2 + 5 = 7 → 2 desirable sums (it should be obvious that the next three will work)
3 + 3 = 6; 3 + 4 = 7; 3 + 5 = 8 → 3 desirable sums
6 desirable sums out of 9 possibilities is
\(\frac{6}{9} = \frac{2}{3}\)
b) 1 * 3 = 3; 1 * 4 = 4; 1 * 5 = 5 → 0 desirable products
2 * 3 = 6; 2 * 4 = 8; 2 * 5 = 10 → 3 desirable products (it should be obvious that the next three will work)
3 * 3 = 9; 3 * 4 = 12; 3 * 5 = 15 → 3 desirable products
6 desirable products out of 9 possibilities is
\(\frac{6}{9} = \frac{2}{3}\)
The worst thing you can do when you see a question is seize up and become afraid to make a move. Even a step in the right direction can get things going sometimes, and an elementary way of solving a question is just as valid as any other way.
Good luck with your studies.
- Andrew
