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17 Jul 2024, 07:00
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C
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Difficulty:
95%
(hard)
Question Stats:
50% (02:09) correct
50%
(02:26)
wrong
based on 341
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If 6 different colored marbles are randomly split into two groups such that each group has at least one marble, what is the probability that the groups consist of an equal number of marbles?
A. 5/31
B. 10/41
C. 10/31
D. 20/41
E. 20/31
A. 5/31
B. 10/41
C. 10/31
D. 20/41
E. 20/31
17 Jul 2024, 08:30
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Bunuel
GMAT Club Official Explanation:
The number of ways to split the marbles into 1 marble in one group and the remaining 5 in another is 6C1 = 6.
The number of ways to split the marbles into 2 marbles in one group and the remaining 4 in another is 6C2 = 15.
The number of ways to split the marbles equally is 6C3/2 = 10. We divide by 2 because 6C3 will produce duplicate splits. For example, one of the triplets given by 6C3 will be {1, 2, 3} and there will also be {4, 5, 6}. However, choosing {1, 2, 3} for one group would mean that the other group is {4, 5, 6}, and similarly choosing {4, 5, 6} for one group would mean that the other group is {1, 2, 3}. Thus, we'd get the same split: {1, 2, 3} - {4, 5, 6} and {4, 5, 6} - {1, 2, 3}. This means that 6C3 will have twice the number of actual splits possible.
Therefore, the probability is 10/(6 + 15 + 10) = 10 / 31.
Answer: C.
General Discussion
jack5397
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17 Jul 2024, 07:13
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IMO - C
Ways to distribute chocolates among 2 group such that each group has at least one.
(1,5) (2,4) (3,3) (4,2) (5,1)
Now only (3,3) has equal marbles. So now selecting 3 marbles for group 1. 3C6=20
6C1 + 6C2+ 6C3 + 6C4 + 6C5 = 6 + 15 + 20 + 15 + 6 = 62
Now probability = 20/62 = 10/31. Hence C.
Ways to distribute chocolates among 2 group such that each group has at least one.
(1,5) (2,4) (3,3) (4,2) (5,1)
Now only (3,3) has equal marbles. So now selecting 3 marbles for group 1. 3C6=20
6C1 + 6C2+ 6C3 + 6C4 + 6C5 = 6 + 15 + 20 + 15 + 6 = 62
Now probability = 20/62 = 10/31. Hence C.
