Mathsbuddy wrote:p1(blue) = 5/12
p2(green) = 4/11
p3(yellow) = 3/10

P(all 3) = 5/12 x 4/11 x 3/10 = 60/1320 = 1/22

REMEMBER: we MULTIPLY probabilities for AND logic

GMATinsight wrote:
The answer Option D is INCORRECT

The first method 5/12 * 4/11 * 3/10 = 1/22 requires multiplication of 3! in numerator as the order of colors need to be taken into account {BGY, BYG, GBY, GYB, YBG, YGB} which can be arranged in 3! ways therefore alternate correct method would be

5/12 * 4/11 * 3/10 * 3! = 6/22 = 3/11

GMATGuruNY wrote:

GMATinsight wrote:
The answer Option D is INCORRECT

The first method 5/12 * 4/11 * 3/10 = 1/22 requires multiplication of 3! in numerator as the order of colors need to be taken into account {BGY, BYG, GBY, GYB, YBG, YGB} which can be arranged in 3! ways therefore alternate correct method would be

5/12 * 4/11 * 3/10 * 3! = 6/22 = 3/11

There is no need here to multiply by 3!.
The question stem asks the following:
What is the probability of drawing a blue ball, green ball and a yellow ball in that order?
The phrase in red implies that the only favorable ordering is BGY.
Thus:
From 5 red marbles, 4 green marbles, and 3 red marbles, P(BGY) = 5/12 * 4/11 * 3/10 = 1/22.

The correct answer is D.

nikhilgmat31 wrote:

GMATGuruNY wrote:

GMATinsight wrote:


What the term specifies here. Do we need to use formula for Permutation.

Please suggest.

We will not use a permutation here because of the specific order.

I usually think of permutations and combinations in terms of spots and options for each spot, this way you can eliminate the formula.

5 x 4 x 3 for the numerator

12 x 11 x 10 for the denominator.

After some cancellation 1/22