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

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Senior Manager
Status: 1,750 Q's attempted and counting
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A box contains three red marbles and 1 green marble. If the  [#permalink]

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Updated on: 11 Oct 2013, 01:39
00:00

Difficulty:

15% (low)

Question Stats:

76% (01:12) correct 24% (01:15) wrong based on 158 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, 13:49.
Last edited by Bunuel on 11 Oct 2013, 01:39, 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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10 Oct 2013, 20:20
1
It is 3/4*2/3*1/2 = 1/4
Senior Manager
Status: 1,750 Q's attempted and counting
Affiliations: University of Florida
Joined: 09 Jul 2013
Posts: 496
Location: United States (FL)
Schools: UFL (A)
GMAT 1: 600 Q45 V29
GMAT 2: 590 Q35 V35
GMAT 3: 570 Q42 V28
GMAT 4: 610 Q44 V30
GPA: 3.45
WE: Accounting (Accounting)
Re: A box contains three red marbles and 1 green marble.  [#permalink]

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22 Oct 2013, 07:43
2
Official Explanation -

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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31 Aug 2018, 05:57
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, 05:57
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