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

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

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07 Aug 2011, 10:56
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Question Stats:

81% (01:15) correct 19% (02:25) wrong based on 36 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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Senior Manager
Joined: 24 Mar 2011
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Re: PS - 700 level - marbles [#permalink]

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07 Aug 2011, 11:02
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, 13:59
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, 19:35
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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Manager
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 10:03
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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Current Student
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Re: PS - 700 level - marbles [#permalink]

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09 Aug 2011, 13:33
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, 14:50
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, 14:54
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, 15:08
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, 15:11
Just take the square root of both sides...

Actually squaring that out algebraicly is terribly unnecessary.

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

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07 Oct 2017, 18:25
You can solve this problem algebraically but if you use intuition a little it'll be much faster.

First of all, we know this question is structured to have "replacement". That basically means we put marbles back in the bag once we draw them. This means the probability of drawing one red marble (then putting it back) and drawing another red marble is the same.

With that said, we know the probability of drawing 2 NON-RED marbles is 16/25. This means drawing 1 NON-RED marble is 4/5. Then we know the probability of drawing a RED marble is 1-4/5 = 1/5.

Then by definition, Prob(red marble) = #RED/#TOTAL = 1/5 Thus 12/Total = 1/5 => Total = 60!
_________________

Insanity at its finest.

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Re: A bag contains 12 red marbles. If someone were to remove 2   [#permalink] 07 Oct 2017, 18:25
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