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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
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Thanks Engr2012. You are correct. I edited the question to reflect your comment.

There's a simpler way to solve the problem without using combinations.
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A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
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Think of it this way: there are initially 12 marbles that can be selected, 4 of each color. So, the odds of selecting any one color on the first pull is 1/3. Say that you select a Red marble, the odds of selecting a Red marble on the second pull is 3/11 (3 red marbles left out of the 11 marbles). The third pull then leaves you with the probability of 2/10 or 1/5 in selecting a Red marble. Finally, on the last pull, you have a 1/9 probability of selecting the last Red marble. Now, the question is reall yasking what the probability of selecting the same color marble on each of the first 4 selections is. That would be expressed as 1/3 x 3/11 x 1/5 x 1/9. Factoring the 3's out of the numerator and denominator leaves us with 1/11 x 1/5 x/ 1/9, or 1/495. Answer A is correct.

However, since there are 3 separate colors and thus, 3 ways to achieve the result, the correct answer is 3 x 1/495, or 1/165. I stand corrected.

*** REVISION - A better way to think of this is that, really, the first selection does not even matter. You WILL select a marble - thus, the Probability is 1/1. What mattters is that the following three selections all return the same color as the 1st. As a result, we have 1/1 x 3/11 x 2/10 x 1/9, or 3/11 x 1/5 x 1/9 = 3/495 0r 1/165.

Originally posted by VeritasPrepDennis on 16 Jul 2015, 12:18.
Last edited by VeritasPrepDennis on 17 Jul 2015, 09:08, edited 4 times in total.
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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
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I agree that the chance of selecting one color four times in a row is 1/495. However, couldn't this happen in 3 different ways, since there are three colors? That would make the answer 3 * 1/495 = 3/495 = 1/165.

In other words, you can select any color marble on the first attempt. Then, you have to select the same color marble on the next three attempts. This would make the calculation

1/1 * 3/11 * 2/10 * 1/9 = 1/165
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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
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Jar contains {4-red, 4-green, 4-blue} marbles. We are asked to find the probability that after removing 4 marbles only 2 colours remain in the jar, i.e., the probability of removing EITHER red, OR green, OR blue.

Sample space = 12C4 = 495.

Probability of choosing 4 red = 4C4/12C4 = 1/495.
Probability of choosing 4 green = 4C4/12C4 = 1/495.
Probability of choosing 4 blue = 4C4/12C4 = 1/495.

Required probability = 1+1+1/495 = 3/495 = 1/165. Ans (B).
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A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
TheGmatTutor wrote:
Thanks Engr2012. You are correct. I edited the question to reflect your comment.

There's a simpler way to solve the problem without using combinations.


Sure. But I feel it is always straightforward if I stick to first principles for Probability, Permutations and combinations. This way , scope of error reduces atleast for me.

Good question though based on the % of correct responses.
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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
Its answer D

Statement 1: Find formula for 5/12 probability

b=black marbles, w=white marbles, T=total marbles

b/T * b-1/T-1 = 5 /12 Trial error: 2/3*1/2 /= 5/12 .... 6/9*5/8= 30/72 = 5/12 --> sufficient

Statement 2: Start with 1/4

w/T=1/3 and w-1/T-1 = 1/4 --> solve for w --> sufficient

Time: 2:30 min
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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
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TheGmatTutor wrote:
A jar contains 12 marbles consisting of an equal number of red, green, and blue marbles. Four marbles are removed from the jar and discarded. What is the probability that only two colors will remain in the jar after the four marbles have been removed?

(A) 1/495
(B) 1/165
(C) 1/81
(D) 1/3
(E) 1/2



There are 4 red, 4 green, and 4 blue marbles in the jar.

If two colors are to remain in the jar after 4 are removed, all 4 marbles removed must be of the same color, that is, they are all red, or all green, or all blue.

Since there are equal number of each color, we can determine the probability of getting all marbles of one color removed and then multiply by 3 (because there are 3 colors of marbles).

The number of ways to get all red marbles is:

4C4 = 1

The total number of ways to select 4 marbles from 12 is:

12C4 = 12!/[4!(12-4)!] = 12!/(4!8!) = (12 x 11 x 10 x 9)/(4 x 3 x 2) = (11 x 5 x 9) = 495

Thus, the probability that all red marbles are removed is 4C4/12C4 = 1/495. However, since there are 3 ways to get all marbles of the same color, the the probability that all same-colored marbles are removed is is 1/495 x 3 = 3/495 = 1/165.

Answer: B
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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
TheGmatTutor wrote:
A jar contains 12 marbles consisting of an equal number of red, green, and blue marbles. Four marbles are removed from the jar and discarded. What is the probability that only two colors will remain in the jar after the four marbles have been removed?

(A) 1/495
(B) 1/165
(C) 1/81
(D) 1/3
(E) 1/2

Source: Original question


Total outcomes of selecting a pack of 4 balls out of 12 = 12C4=495

If a pack of four balls of any single colour is selected,then only two colors remain. Therefore total desired outcomes of getting packs of 4 balls of same colour=3( packs of 4 balls of same colour)

Therefore required probability = 3/495= 1/165
Answer B

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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
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Re: A jar contains 12 marbles consisting of an equal number of red, green, [#permalink]
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