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
The OA is B.
Please, can any expert explain this PS question for me? I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.
A jar contains 12 marbles consisting of an equal...
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We're starting with: R, R, R, R, G, G, G, G, B, B, B, Bswerve 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, P(only 2 colors remaining) = P(all 4 selected marbles are the SAME color)
P(all 4 selected marbles are the SAME color) = P(1st marble is ANY color AND 2nd marble matches 1st marble AND 3rd marble matches 1st marble AND 4th marble matches 1st marble)
= P(1st marble is ANY color) x P(2nd marble matches 1st marble) x P(3rd marble matches 1st marble) x P(4th marble matches 1st marble)
= 1 x 3/11 x 2/10 x 1/9
= 1/165
Answer: B
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- Scott@TargetTestPrep
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There are 4 red, 4 green, and 4 blue marbles in the jar.swerve 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
If two colors are to remain in the jar after 4 are removed, either all red, all green, or all blue marbles must be removed.
Since there are equal number of each color, we can determine the probability of getting all of one color removed and then multiply by 3.
The number of ways to get all red:
4C4 = 1
The total number of ways to select the marbles 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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GMATH´s method suggests avoiding the "partial probabilities sequences" (correctly used in previous post) because:swerve 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
1. The justification for its validity is out-of-GMAT´s scope (conditional probabilities, NOT independency).
2. There are some problems in which students usually fall into "traps".
All that mentioned, let´s see how we would deal with this problem!
\[4\,{\text{red}}\,\,,\,\,4\,\,{\text{green,}}\,\,{\text{4}}\,\,{\text{blue}}\,\,\,\, \to \,\,\,\,{\text{4}}\,\,{\text{taken}}\,\,{\text{out}}\]
\[? = P\left( {{\text{only}}\,\,{\text{2}}\,\,colors\,\,{\text{remain}}} \right)\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{? = P\left( {4\,\,{\text{taken}}\,\,{\text{out}}\,\,{\text{same}}\,\,{\text{color}}} \right)}\]
\[{\text{total}} = C\left( {12,4} \right)\,\,\,\,\left( {{\text{equiprobables}}} \right)\]
\[{\text{favorable}}\,\,{\text{ = }}\,\,3\,\,\,\,\,\,\left( {{\text{all}}\,\,{\text{red}}\,\,{\text{or}}\,\,{\text{all}}\,\,{\text{green}}\,\,{\text{or}}\,\,{\text{all}}\,\,{\text{blue}}} \right)\]
\[? = \frac{3}{{C\left( {12,4} \right)}} = \frac{{3\, \cdot 4!}}{{12 \cdot 11 \cdot 10 \cdot 9}} = \ldots = \frac{1}{{165}}\]
The above follows the notations and rationale taught in the GMATH method.
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