A bag contains 12 red marbles. If someone were to remove 2 : GMAT Problem Solving (PS)
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# A bag contains 12 red marbles. If someone were to remove 2

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A bag contains 12 red marbles. If someone were to remove 2 [#permalink]

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07 Aug 2011, 09:56
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45% (medium)

Question Stats:

65% (01:51) correct 35% (01:25) wrong based on 17 sessions

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A bag contains 12 red marbles. If someone were to remove 2 marbles from the bag, one at a time, and replace the first marble after it was removed, the probability that neither marble would be red is 16/25. How many marbles are in the bag?

A. 24
B. 48
C. 60
D. 72
E. 84
[Reveal] Spoiler: OA
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Re: PS - 700 level - marbles [#permalink]

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07 Aug 2011, 10:02
bschool83 wrote:
A bag contains 12 red marbles. If someone were to remove 2 marbles from the bag, one at a time, and replace the first marble after it was removed, the probability that neither marble would be red is 16/25. How many marbles are in the bag?

24
48
60
72
84

Answer C. let x be total marbles. hence x-12 are other color marbles.
now $$(x-12)/x * (x-12)/x = 16/25$$
solving this for x, x = 60
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Re: PS - 700 level - marbles [#permalink]

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07 Aug 2011, 12:59
piyatiwari wrote:
agdimple333 wrote:
now $$(x-12)/x * (x-12)/x = 16/25$$

probability of non-red marble drawn = (x-12)/x
since the marble is placed back, probability of non-red marble drawn again = (x-12)/x

so, (x-12)/x * (x-12)/x = 16/25

(x-12)^2 = x^2 * 16/25

==> (x-12) = x * 4/5

x/5 = 12
==> x = 60.
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Re: PS - 700 level - marbles [#permalink]

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07 Aug 2011, 13:04
Got it. Thanks agdimple333!
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Re: PS - 700 level - marbles [#permalink]

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07 Aug 2011, 18:35
agdimple333 wrote:
piyatiwari wrote:
agdimple333 wrote:
now $$(x-12)/x * (x-12)/x = 16/25$$

probability of non-red marble drawn = (x-12)/x
since the marble is placed back, probability of non-red marble drawn again = (x-12)/x

so, (x-12)/x * (x-12)/x = 16/25

(x-12)^2 = x^2 * 16/25

==> (x-12) = x * 4/5

x/5 = 12
==> x = 60.

How did that happen? I understand where you get (x-12)^2 and x^2 but how did x^2 get on the other side of the =?

Understand the square root to remove the exponents and 16/25 for the final answer just missing that little middle step....
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 09:03
ok let me see if i can explain what went on in the previous post

lets say i have x marbles in the bag in total --> out of them 12 are red

so the probability of pulling a non-red marble is (x -12) / x

now the marble is placed back in the bag and we have x marbles again, of which again 12 are red. so the probability of pulling a non-red marble out is (x-12) / x

probability theorm states that if the probability of event A occuring is m and the probability of event B occuring is n then the probability of both A and B occuring is m*n

so therefore the probability of 2 non-red marbles getting pulled out is [(x-12)/x ] * [(x-12)/x]

this is given as 16/25

--> (x-12)^2 = 16/25
x^2

square rooting u have x-12/x = 4/5 or x = 60.

hope that helps
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 12:33
My much simpler solution...

(n-12)/n chance of getting a marble that isn't red. Since we do it twice with replacement we just square that.

[(n-12)/n]^2 = 16/25
(n-12)/n = 4/5
5n-60 = 4n
n = 60
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 13:50
far257 wrote:
My much simpler solution...

(n-12)/n chance of getting a marble that isn't red. Since we do it twice with replacement we just square that.

[(n-12)/n]^2 = 16/25
(n-12)/n = 4/5
5n-60 = 4n
n = 60

could you care to explain the math behind the first part: [(n-12)/n]^2
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 13:54
n marbles of which 12 are not red. So chance of getting a not-red marble is n-12/n. Since we have replacement we just do it twice, so we just square it.

Like what's the chance of getting heads on a coin twice? 1/2*1/2. or (1/2)^2.
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 14:08
far257 wrote:
n marbles of which 12 are not red. So chance of getting a not-red marble is n-12/n. Since we have replacement we just do it twice, so we just square it.

Like what's the chance of getting heads on a coin twice? 1/2*1/2. or (1/2)^2.

I meant could you show the steps of the squaring...whether you multiply it twice or square it...i'm not getting 16/25. Thanks!
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 14:11
Just take the square root of both sides...

Actually squaring that out algebraicly is terribly unnecessary.
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 14:43
Got it!!! sheesh this one took forever. thanks!
Re: PS - 700 level - marbles   [#permalink] 09 Aug 2011, 14:43
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