GMAT Club Olympics 2024 (Day 8): If 6 different colored marbles are

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. 5/31
B. 10/41
C. 10/31
D. 20/41
E. 20/31

jack5397 · Accepted Answer

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.

Fido10 · Answer

We have 5 ways to select the marbles, 
Wether one groupe contains 1 marble and the other 5 and number of ways to chose this is : 6C1 * 5C5 =6
Wether one groupe contains 2 marble and the other 4 and number of ways to chose this is : 6C2 * 4C4 =15
 Wether one groupe contains 3 marble and the other 3 and number of ways to chose this is : 6C3 * 3C3 = 20 
Wether one groupe contains 4 marble and the other 2 and number of ways to chose this is : 6C4 * 2C2 = 15
Wether one groupe contains 5 marble and the other 1 and number of ways to chose this is : 6C5 * 1C1 =6

the total number of possibilities then is 62

The number of ways, then,the two groups will have the same number of marbles is = 20/62 =  10/31 
 Correct answer is C

Bunuel · Answer

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.

Dooperman · Answer

Choose the number of marbles for 1st group and the number of marbles for the other group would be decided automatically. 
Total Choices = 6C0 + 6C1 + 6C2+ 6C3 + 6C4 + 6C5 + 6C6. Out of this we have to subtract 6C0 and 6C6 as both the groups should have at least 1 marble. 
So total choices = 62.

Ways of selecting 3 marbles for 1st group = 6C3 = 20.

P= 20/62 = 10/31.

BKD5050 · Answer

why 6c1 and 6c2 is not divided by 2!? just like 6c3

Bunuel · Answer

The reason 6C1 and 6C2 are not divided by 2! is that these splits don't involve symmetry. For example, when selecting 1 marble for one group (6C1), there’s only one way to assign the remaining marbles to the other group. Similarly, with 6C2, the two groups are not interchangeable.

However, in the case of splitting 6 marbles into two equal groups (6C3), the two groups are indistinguishable, so we divide by 2! to account for this duplication (since the groups are interchangeable).

GMAT Club Olympics 2024 (Day 8): If 6 different colored marbles are

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