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by Mathsbuddy » Thu Nov 27, 2014 11:53 am
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
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by GMATinsight » Thu Nov 27, 2014 7:35 pm
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
You are right about the logic of MULTIPLY for AND

Another point to note here is "Because the order is given BYG therefore selection method would require the denominator to be multiplied by 3! as in total outcomes, all the orders must be taken into account"

The correct answer of the question should be
(5C1 x 4C1 x 3C1)/12C3x3! = 60 / 220X6 = 6/22 = 1/22

The answer Option D is CORRECT

The first method 5/12 * 4/11 * 3/10 = 1/22 Doesn't require multiplication of 3! in numerator only because ordering of colors is fixed and the order of colors does NOT 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 = 1/22
Last edited by GMATinsight on Fri Nov 28, 2014 5:51 am, edited 1 time in total.
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by GMATGuruNY » Fri Nov 28, 2014 3:32 am
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.
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by GMATinsight » Fri Nov 28, 2014 5:52 am
Careless reading and understanding question from others' explanation of the question is too dangerous!!!
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by nikhilgmat31 » Mon Jul 20, 2015 4:12 am
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.
I got it correct But Even If we follow other order, let's say G B Y then the probability is also same.

G = 4/12
B = 5/11
Y = 3/10

G*B*Y = 4/12 * 5/11 * 3/10 = 1/22

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

Please suggest.
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by Jim@StratusPrep » Mon Jul 20, 2015 5:10 am
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
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by nikhilgmat31 » Wed Jul 29, 2015 10:09 pm
Please suggest when to use nCr formula to select a item for a group.

lets says we need to select 1 blue ball from 5 blue,4 Green & 3 Yellow Balls. - then we can simple take 5/12
Then What is the need of writing 5C1/12 - however 5C1 also gives 5

In other case

lets says we need to select 2 blue ball from 5 blue,4 Green & 3 Yellow Balls. - then we can simple take 5/12 * 4/11 = 5/33

In that case do we need to write it as 5C2 = 10

What will be denominator in this case
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