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A box contains three red marbles and 1 green marble. If the

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Director
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A box contains three red marbles and 1 green marble. If the  [#permalink]

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New post Updated on: 11 Oct 2013, 02:39
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Question Stats:

76% (00:48) correct 24% (01:00) wrong based on 144 sessions

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A box contains three red marbles and 1 green marble. If the marbles are removed from the box one at a time, in random order, what is the probability that all three red marbles are removed before the green marble?

(A) \(\frac{1}{64}\)

(B) \(\frac{1}{24}\)

(C) \(\frac{1}{12}\)

(D) \(\frac{1}{4}\)

(E) \(\frac{1}{2}\)

Originally posted by avohden on 10 Oct 2013, 14:49.
Last edited by Bunuel on 11 Oct 2013, 02:39, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A box contains three red marbles and 1 green marble.  [#permalink]

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New post 10 Oct 2013, 21:20
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It is 3/4*2/3*1/2 = 1/4
Director
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GMAT 1: 600 Q45 V29
GMAT 2: 590 Q35 V35
GMAT 3: 570 Q42 V28
GMAT 4: 610 Q44 V30
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WE: Accounting (Accounting)
Re: A box contains three red marbles and 1 green marble.  [#permalink]

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New post 22 Oct 2013, 08:43
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Official Explanation -

Answer D
There are 3 red marbles and 4 marbles total, so the probability of drawing a red marble on the first draw is 3/4 . On the second draw there are 2 red marbles and three marbles total so the probability of drawing a red marble is 2/3. On the third draw, there is only one red marble and one green marble, so the probability of drawing the red marble is 1/2. The probability of all three events happening is the product of the probabilities,\(\frac{3}{4}* \frac{2}{3} *\frac{1}{2}or \frac{1}{4}\).

Hope it helps.
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Re: A box contains three red marbles and 1 green marble. If the  [#permalink]

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New post 31 Aug 2018, 06:57
avohden wrote:
A box contains three red marbles and 1 green marble. If the marbles are removed from the box one at a time, in random order, what is the probability that all three red marbles are removed before the green marble?

(A) \(\frac{1}{64}\)

(B) \(\frac{1}{24}\)

(C) \(\frac{1}{12}\)

(D) \(\frac{1}{4}\)

(E) \(\frac{1}{2}\)


GMATH´s method suggests avoiding the "partial probabilities sequences" (correctly used in the previous posts) because:

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!

\(3\,{\text{red}}\,\,,\,\,1\,\,{\text{green}}\)

\(? = P\left( {{\text{green}}\,\,{\text{last,}}\,\,{\text{in}}\,\,{\text{4}}\,\,{\text{sequential}}\,\,{\text{extractions}}} \right)\)

\({\text{total}} = 4!\,\,\,\,\left( {{\text{equiprobables}}} \right)\)

\({\text{favorable}}\,\,{\text{ = }}\,\,{\text{3!}}\,\,\,\,\,\,\left( {{P_3}\,\,{\text{in}}\,\,{\text{first}}\,\,{\text{3}}\,\,{\text{red}}\,\,{\text{extractions}}} \right)\)

\(? = \frac{{3!}}{{4!}} = \frac{1}{4}\)

The above follows the notations and rationale taught in the GMATH method.
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Re: A box contains three red marbles and 1 green marble. If the &nbs [#permalink] 31 Aug 2018, 06:57
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