GMAT Problem Solving- An Urn Contains 6 Red, 4 Blue, 2 Green and 3 Yellow marbles

Question- An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?

  1. 13/35
  2. 24/455
  3. 11/15
  4. 1/13
  5. 2/455

“An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

It is asked in the question to find what is the probability that out of all marbles 1 is green, 2 are blue and 1 is red?
Let us evaluate further
Total marbles in urn = 6+4+2+3 which is 15..
The Methods to choose 4 of them = 15C4
= \(\frac{15*14*13*12}{4*3*2}\)
= 15 * 7 *13
Same way,
to choose one green, two blues, and one red = 2C1*4C2*6C1
=2*6*6
According to probability
\(\frac{2*6*6}{15*7*13}\)
= \(\frac{24}{455}\)
The answer is 24/455

Correct Answer: C

Model Answer 2

There is another approach to solve this question which is pretty easy
It is given
Total marbles: (6 + 4 + 2 + 3), which is 15.
E shall be the drawing of one green, two blue, and one red marbles.
Then,
n (E) = 2C1*4C2*6C1
= 2 * \(\frac{4*3}{2*1}\) * 6
=72
So n (S) = 15C4
= \(\frac{15*14*13*12}{4*3*2}\)
=1365
Therefore probability will be
P (E) = \(​​\frac{n(E)}{n(S)}\)
=\(\frac{72}{1365}\)
= \(\frac{24}{455}\)
The answer is 24/455

Correct Answer: C

Approach Solution 3:

The inquiry seeks to determine what the likelihood is that out of all the marbles, one is green, two are blue, and one is red.
Let's continue to evaluate.
Total marbles in the urn are 15 (6 + 4 + 2 + 3).
The formulas to select four of these are: 15C4

= \(\frac{15*14*13*12}{4*3*2}\)
= 15 * 7 *13

The methods to select 4 of them are as follows: 15C4 = = 15 * 7 *13
Similarly, to select one green, two blues, and one red, use the formula 2C1*4C2*6C1 = 2*6*6.
Probabilistically speaking.

Correct Answer: C

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