A jar contains 12 marbles consisting of an equal number of

Jay@ManhattanReview wrote:

BTGmoderatorDC 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

So the situation is that there are 4 marbles each of red, green, and blue colors.

We have to find out the probability that only two colors will remain in the jar after the four marbles have been removed and discarded.

Say the 4 marbles removed are red, thus,

the probability that only green, and blue colors will remain in the jar after 4 red marbles have been removed = 4C4 / 12C4 = 1/12C4 = (1.2.3.4)/(12.11.10.9) = 1/(55.9)

In the same way, the 4 marbles removed can be green or blue, thus,

the probability that only two colors will remain in the jar after the four marbles have been removed and discarded
= 3*[1/(55.9)] = 1/165

The correct answer: B

Hope this helps!

-Jay
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55.9 should be 495

BTGmoderatorDC 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

For only 2 colors to remain after 4 marbles have been removed, the 4 selected marbles must be of the same color.
The first selected marble can be of ANY COLOR.
A good outcome will be yielded if the second, third and fourth marbles are of the SAME COLOR as the first.
P(2nd marble is of the same color as the first) = 3/11. (Of the 11 marbles that remain after the removal of the 1st marble, 3 are of the same color as the first.)
P(3rd marble is of the same color as the first 2) = 2/10. (Of the 10 remaining marbles, 2 are of the same color as the first two.)
P(4th marble is of the same color as the first 3) = 1/9. (Of the 9 remaining marbles, 1 is of the same color as the first three.)
To combine these probabilities, we multiply:
3/11 * 2/10 * 1/9 = 1/165.

The correct answer is B.

GMATGuruNY wrote:

BTGmoderatorDC 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

For only 2 colors to remain after 4 marbles have been removed, the 4 selected marbles must be of the same color.
The first selected marble can be of ANY COLOR.
A good outcome will be yielded if the second, third and fourth marbles are of the SAME COLOR as the first.
P(2nd marble is of the same color as the first) = 3/11. (Of the 11 marbles that remain after the removal of the 1st marble, 3 are of the same color as the first.)
P(3rd marble is of the same color as the first 2) = 2/10. (Of the 10 remaining marbles, 2 are of the same color as the first two.)
P(4th marble is of the same color as the first 3) = 1/9. (Of the 9 remaining marbles, 1 is of the same color as the first three.)
To combine these probabilities, we multiply:
1/11 * 2/10 * 1/9 = 1/165.

The correct answer is B.

My day for nagging corrections I guess. 1/11 should be 3/11

regor60 wrote:My day for nagging corrections I guess. 1/11 should be 3/11

Fixed!

BTGmoderatorDC 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