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29 Jul 2024, 04:32
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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. \(\frac{5}{31}\)
B. \(\frac{10}{41}\)
C. \(\frac{10}{31}\)
D. \(\frac{20}{41}\)
E. \(\frac{20}{31}\)
A. \(\frac{5}{31}\)
B. \(\frac{10}{41}\)
C. \(\frac{10}{31}\)
D. \(\frac{20}{41}\)
E. \(\frac{20}{31}\)
29 Jul 2024, 04:32
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Official Solution:
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. \(\frac{5}{31}\)
B. \(\frac{10}{41}\)
C. \(\frac{10}{31}\)
D. \(\frac{20}{41}\)
E. \(\frac{20}{31}\)
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 \(\frac{6C3}{2} = 10\). 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 \(\frac{10}{6 + 15 + 10} = \frac{10}{31}\).
Answer: C
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. \(\frac{5}{31}\)
B. \(\frac{10}{41}\)
C. \(\frac{10}{31}\)
D. \(\frac{20}{41}\)
E. \(\frac{20}{31}\)
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 \(\frac{6C3}{2} = 10\). 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 \(\frac{10}{6 + 15 + 10} = \frac{10}{31}\).
Answer: C
26 Sep 2024, 13:49
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Hey Bunuel
If we take two groups namely A and B, then shouldn't be there 5 cases {(1,5), (2,3), (3,3), (4,2), (5,1)} and answer should be 20/62?
If we take two groups namely A and B, then shouldn't be there 5 cases {(1,5), (2,3), (3,3), (4,2), (5,1)} and answer should be 20/62?
