12 Marbles are Selected at Random from a Large Collection of White, Red, Green and Yellow Marbles GMAT Problem Solving

Question: 12 marbles are selected at random from a large collection of white, red, green and yellow Marbles. The number of Marbles of each colour is unlimited. Find the probability that the selection contains at least one marble of each colour?

1. 34/91
2. 23/31
3. 36/91
4. 33/91
5. ¼

“12 marbles are selected at random from a large collection of white, red, green and yellow Marbles. The number of Marbles of each colour is unlimited. Find the probability that the selection contains at least one marble of each colour?” - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Quantitative Review". GMAT Quant section consists of a total of 31 questions. To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical understanding.

Approach Solution 1:

The above-given problem can be solved using integer rules. Let us consider, W+R+G+Y=12, where all are positive integers.

Case I:

It is given that first, it gives one to each, so 4 are gone. Now, distributing the remaining 8 to anyone as the restriction of at least one is done.

W+R+G+Y=8.
W+R+G+Y=8.
Now, take this as 8 marbles and 3 partitions, so total 11 things. You can choose these 3 partitions in 11C3 ways

\(\frac{11!}{8!3!}=\frac{11*10*9}{3*2}=11*5*3\)

Case II:

In another case, considering W+R+G+Y=12. Now, take this as 12 marbles and 3 partitions, so total 15 things. You can choose these 3 partitions in 15C3 ways=

\(\frac{15!}{12!3!}=\frac{15*14*13}{3*2}=5*7*13\)

Probability = \(\frac{11*5*3}{5*7*13}=\frac{33}{91}\)

Hence, the probability that the selection contains at least one marble of each color is \(\frac{33}{91}\)

Correct Answer: D

Approach Solution 2:

The above-given problem can be solved using set rules. Let us consider that there are 3 lines or partitions and 12 marbles. These 3 partitions divide the 12 marbles in 4 groups (W, R, G and Y).

Considering Case 1:

When there is at least 1 marble in each group.
Total number of ways= n-1Cr-1= 11C3 {n=12 and r=4}

Considering Case 2:

When there are no restrictions.
Total number of ways= n+r-1Cr-1= 15C3
Probability= 11C3/15C3= 11*10*9/15*14*13= 33/91
Hence, the probability that the selection contains at least one marble of each color is \(\frac{33}{91}\)

Correct Answer: D

Approach Solution 3:

Let us check the total cases:

Case 1: x1+x2+x3+x4=12…C(n+r−1,r−1)

=(12+4−1,4−1)

=15!3!12!x1+x2+x3+x4

=12…C(n+r−1,r−1)

=(12+4−1,4−1)

=15!3!12!

We can see the favorable cases be:

​Case 2: x1+1+x2+1+x3+1+x4+1

=12…x1+x2+x3+x4

=12−4

=8x1+1+x2+1+x3+1+x4+1

=12…x1+x2+x3+x4

=12−4=8

We get,

C(n+r−1,r−1)

=(8+4−1,4−1)

=11!3!8!C(n+r−1,r−1)

=(8+4−1,4−1)

 =11!3!8!

Probability: Favorable/Total

11!3!8!15!3!12!

=11!(3!12!)3!8!(15!)

=11!12!8!15!=339111!3!8!15!3!12!

=11!(3!12!)3!8!(15!)

=11!12!8!15!=3391

Correct Answer: D

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