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GRE Prep Club Team Member
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A certain bag of gemstones is composed of two-thirds diamonds and one-
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04 May 2021, 22:01
04 May 2021, 22:01
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A certain bag of gemstones is composed of two-thirds diamonds and one-third rubies. If the probability of randomly selecting two diamonds from the bag, without replacement, is 5/12, what is the probability of selecting two rubies from the bag, without replacement?
(A) 5/36
(B) 5/24
(C) 1/12
(D) 1/6
(E) 1/4
(A) 5/36
(B) 5/24
(C) 1/12
(D) 1/6
(E) 1/4
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Re: A certain bag of gemstones is composed of two-thirds diamonds and one-
[#permalink]
05 May 2021, 11:04
05 May 2021, 11:04
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GeminiHeat wrote:
A certain bag of gemstones is composed of two-thirds diamonds and one-third rubies. If the probability of randomly selecting two diamonds from the bag, without replacement, is 5/12, what is the probability of selecting two rubies from the bag, without replacement?
(A) 5/36
(B) 5/24
(C) 1/12
(D) 1/6
(E) 1/4
(A) 5/36
(B) 5/24
(C) 1/12
(D) 1/6
(E) 1/4
Let the total number of gems be \(x\)
D = \(\frac{2x}{3}\)
R = \(\frac{x}{3}\)
Given: \(\frac{D_1}{x} . \frac{D_2}{x-1} = 5/12\)
\(D_1 = \frac{2x}{3}\)
\(D_2 = \frac{2x}{3} - 1 = \frac{2x-3}{3}\)
Therefore, \(\frac{2(2x-3)}{9(x-1)} = \frac{5}{12}\)
Solve for \(x = 9\)
Now, Prob. of selecting two Rubies without replacement = \((\frac{3}{9})(\frac{2}{8}) = \frac{1}{12}\)
Hence, option C
gmatclubot
Re: A certain bag of gemstones is composed of two-thirds diamonds and one- [#permalink]
05 May 2021, 11:04
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