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GRE Math Challenge #86- A box contains 20 marbles all of

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GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 09 May 2015, 11:01
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Question Stats:

52% (00:38) correct 47% (00:22) wrong based on 19 sessions
A box contains 20 marbles all of which are solid colored; 5 of the marbles are green and 10 of the marbles are red.

Quantity A
Quantity B
The probability that a marble selected at random form the box will be green
The probability that a marble selected at random from the box will be neither red nor green


• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
[Reveal] Spoiler: OA

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Last edited by sandy on 22 May 2018, 14:25, edited 4 times in total.
Edited by Carcass
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Re: GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 18 Aug 2016, 06:17
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sandy wrote:
A box contains 20 marbles all of which are solid colored; 5 of the marbles are green and 10 of the marbles are fed.

Quantity A: The probability that a marble selected at random form the box will be green
Quantity B: The probability that a marble selected at random from the box will be neither red now green
• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined


in this for quantity A, the probability will be 5/20 = 1/4. when it comes to quantity B, the probability is found out using P(A)*P(B) = (5/20)*(10/20) = 1/8

so the answer is A
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Re: GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 18 Aug 2016, 11:36
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I think both quantities are equal and answer is (C). This is my solution to this problem:

Let probability of selecting a green marble at random from the box is P(G) & probability of selecting a red marble at random from the box is P(R) & probability of selecting a marble at random from the box that is neither nor green is is P(neither G nor R)

P(G)*P(R) = 0, as both are mutually exclusive events.

Quantity A : P(G) = (5/20) =(1/4) ; B : P(neither G nor R) = 1 – (P(G)+P(R)) = 1 – ((5/20)+(10/20)) = 1/4
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Re: GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 05 Dec 2016, 14:01
Answer is C
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Re: GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 15 Dec 2016, 22:26
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I also think C. There are 5 green marbles, 10 red marbles, and that means there must be 5 remaining that are neither red nor green. There is a 5/20 chance that the marble will be green. There is also a 5/20 chance it will be neither red nor green.
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Re: GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 17 Dec 2016, 05:59
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The solution is 'C' : Both quantities are equal

Explanation:
Total number of marbles = 20
Number of green marbles = 5
Number of red marbles = 10
Number of other colored marbles = 5
From a random draw on 20 marbles, probability of picking a green marble = 5/20 = 1/4 = P(G)
From a random draw on 20 marbles, probability of picking an other colored marble = 5/20 = 1/4 = P(OC)
We can see that P(G)=P(OC)=1/4
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Re: GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 22 May 2018, 14:09
Hello Sandy,

I strongly doubt the validity of the answer to this being A. Kindly show how the answer being A is achieved.

sandy wrote:
A box contains 20 marbles all of which are solid colored; 5 of the marbles are green and 10 of the marbles are fed.

Quantity A
Quantity B
The probability that a marble selected at random form the box will be green
The probability that a marble selected at random from the box will be neither red now green


• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
1 KUDOS received
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Re: GRE Math Challenge #86- A box contains 20 marbles all of [#permalink] New post 22 May 2018, 14:26
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Hi,

The answer is indeed C. Fixed it.
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Re: GRE Math Challenge #86- A box contains 20 marbles all of   [#permalink] 22 May 2018, 14:26
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