M43-16

Question

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}

Bunuel · Answer

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

Nipunh · Answer

I did it differently, idk whether it's 100% correct or not please help Bunuel

here's how I did it: Total cases: 6C1 + 6C2 +6C3+ 6C4 + 6C5= 62 Reqd case i.e for getting equal no. of marbles in each group: 6C3=20 hence probability of getting equal no. of marbles = 20/62=10/31 Thanks in advance!

Bunuel · Answer

With your method, you’re actually counting double the total number of cases and double the favorable cases. For example, 6C1 and 6C5 both give the number of ways to split 6 marbles into two groups, one with 1 marble and another with 5. This overcounting balances out, so in the end, you still get the correct answer.

BenTen10 · Answer

Hi  Bunuel ,

I really liked the question and would like to make one small remark.

Do you agree that the question leaves open the total amount of marbles, which go into either group 1 or group 2? Considering that not all marbles would have to be split into the two groups, adds additional scenarios (4C1, 3C1, 2C1,...).

Might make sense to add a note that all 6 marbles have to be split up into the two groups to avoid confusion.

Thanks for your help!

Bunuel · Answer

The question says "6 different colored marbles are randomly split into two groups", which clearly means all 6 marbles are split. I don't see any ambiguity there.

Bhavita. · Answer

I did not quite understand the solution. I solved it this way; not sure why it's wrong..... Since there are 6 different marbles, the number of ways of forming 2 groups is 2^6 (64)... no of ways where all marbles are in either of the groups is 2 .. so no of ways of forming at least one marble is there in either of group is 62

probability that the groups consist of an equal number of marbles = 6c3 / 2

10/62 = 5/31

probability that the groups consist of an equal number of marbles = 6c3 / 2

10/62 = 5/31

Bunuel · Answer

The numerator and denominator must be  consistent  in how the two groups are counted.

In the denominator, 2^6 - 2 = 62 treats the groups as  ordered : putting {1,2,3} in Group 1 and {4,5,6} in Group 2 is different from reversing the two groups.

But in the numerator, 6C3/2 = 10 treats the groups as  unordered , because dividing by 2 identifies those two arrangements as the same split.

So if you use 62 in the denominator, you should use 6C3 = 20 in the numerator.

Probability = 20/62 = 10/31.

M43-16

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