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# A certain bag of gemstones is composed of two-thirds

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Director
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A certain bag of gemstones is composed of two-thirds  [#permalink]

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29 May 2006, 07:32
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66% (02:05) correct 34% (02:08) wrong based on 386 sessions

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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
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Re: A certain bag of gemstones is composed of two-thirds  [#permalink]

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09 Jan 2014, 21:55
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b00gigi wrote:
Can someone explain the

2/3 * (2R-1)/(3R-1)

part?

Say, a bag has 6 diamonds and 3 rubies. What is the probability of selecting 2 diamonds one after the other without replacement?

Probability of selecting one diamond = 6/9
Probability of selecting yet another diamond after selecting one = 5/8 (no of diamonds has gone down by 1 and total no. of diamonds has gone down by 1 too)
Total probability = (6/9)*(5/8)

Here, we assume that no of rubies is R and no of diamonds is 2R (since no of diamonds is twice the no of rubies)
Probability of selecting two diamonds without replacement = (2R/3R) * (2R - 1)/(3R - 1) = 5/12
Either cross multiply to get the value of R or try to plug in some values to see where you get a multiple of 12 in the denominator.
Once you get the value of R as 3, you know the number of diamonds is 6.

Probability of picking two rubies one after the other without replacement = (3/9) *(2/8) = 1/12
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### Most Helpful Community Reply Intern Joined: 05 Apr 2006 Posts: 34 Re: Challenge MHTNGMAT [#permalink] ### Show Tags 29 May 2006, 09:03 3 2 BG 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 is answer 1/12 2/3 * (2X-1) / (3X-1) = 5/ 12 => X = 3 So total gems = 9 and probability of ruby = 1/3 * 2/8 = 1/12 ##### General Discussion VP Joined: 29 Dec 2005 Posts: 1318 Re: Challenge MHTNGMAT [#permalink] ### Show Tags 29 May 2006, 09:46 2 guptaraja wrote: BG 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 is answer 1/12 2/3 * (2X-1) / (3X-1) = 5/ 12 => X = 3 So total gems = 9 and probability of ruby = 1/3 * 2/8 = 1/12 this is good enough......... [{2/3(x)}/x] [{(2x/3)-1}/x-1] = 5/12 x^2-9x = 0 x = 0, 9 so x = 9 diamond = 6 ruby = 3 the prob (2 ruby) = 3c2/9c2 = 1/12 C. Senior Manager Status: Final Lap Joined: 25 Oct 2012 Posts: 262 Concentration: General Management, Entrepreneurship GPA: 3.54 WE: Project Management (Retail Banking) Re: A certain bag of gemstones is composed of two-thirds [#permalink] ### Show Tags Updated on: 09 Feb 2013, 19:23 2 1 BG 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 Let R be the numbers of rubies in the bag, we told that the selection is made without replacement in both cases ( selecting two diamonds or selecting two rubies) Hence, we have : $$\frac{2}{3}*\frac{2R-1}{3R-1}=\frac{5}{12}$$ So, the number of diamonds in the bag is 6. Likewise, the number of rubies in the bag is 3 and the total of the gemstones is 9. The probability of selecting two rubies from the bag without replacement is : $$\frac{1}{3}*\frac{2}{8}=\frac{1}{12}$$ Answer : C _________________ KUDOS is the good manner to help the entire community. "If you don't change your life, your life will change you" Originally posted by Rock750 on 06 Feb 2013, 17:38. Last edited by Rock750 on 09 Feb 2013, 19:23, edited 1 time in total. Manager Joined: 08 Dec 2012 Posts: 64 Location: United Kingdom WE: Engineering (Consulting) Re: A certain bag of gemstones is composed of two-thirds [#permalink] ### Show Tags 09 Feb 2013, 19:13 Rock750 wrote: BG 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 Let R be the numbers of rubies in the bag, we told that the selection is made without replacement in both cases ( selecting two diamonds or selecting two rubies) Hence, we have : $$\frac{2}{3}*\frac{2R-1}{3R-1}=\frac{5}{12}$$ So, the number of diamonds in the bag is 3. Likewise, the number of rubies in the bag is 6 and the total of the gemstones is 9. The probability of selecting two rubies from the bag without replacement is : $$\frac{1}{3}*\frac{2}{8}=\frac{1}{12}$$ Answer : C Looks like you got your diamonds and rubies mixed up though you got it right later Senior Manager Status: Final Lap Joined: 25 Oct 2012 Posts: 262 Concentration: General Management, Entrepreneurship GPA: 3.54 WE: Project Management (Retail Banking) Re: A certain bag of gemstones is composed of two-thirds [#permalink] ### Show Tags 09 Feb 2013, 19:28 nave81 wrote: Rock750 wrote: BG 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 Let R be the numbers of rubies in the bag, we told that the selection is made without replacement in both cases ( selecting two diamonds or selecting two rubies) Hence, we have : $$\frac{2}{3}*\frac{2R-1}{3R-1}=\frac{5}{12}$$ So, the number of diamonds in the bag is 3. Likewise, the number of rubies in the bag is 6 and the total of the gemstones is 9. The probability of selecting two rubies from the bag without replacement is : $$\frac{1}{3}*\frac{2}{8}=\frac{1}{12}$$ Answer : C Looks like you got your diamonds and rubies mixed up though you got it right later u are right navy81, thx Hope this silly mistake had not confused anyone. Anyway, i think it's OK by now _________________ KUDOS is the good manner to help the entire community. "If you don't change your life, your life will change you" VP Status: Far, far away! Joined: 02 Sep 2012 Posts: 1096 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Re: what is the probability of selecting 2 rubees from the bag [#permalink] ### Show Tags 10 Aug 2013, 09:50 4 Diamonds=2x Rubees=x Two D= $$\frac{2x}{3x}*\frac{(2x-1)}{(3x-1)}=\frac{5}{12}$$, solve for x and you get x=0(non sense) or x=3 (better). Diamonds=2*3 Rubees=3 Probability of getting 2 rubees=$$\frac{3}{9}*\frac{2}{8}=\frac{1}{12}$$ _________________ It is beyond a doubt that all our knowledge that begins with experience. Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b] SVP Joined: 06 Sep 2013 Posts: 1851 Concentration: Finance Re: A certain bag of gemstones is composed of two-thirds [#permalink] ### Show Tags 09 Jan 2014, 15:00 4 BG 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 (d/d+r)(d-1/d+r-1) = 5/12 d = 2r Therefore r = 3 d= 6 Probability of 2 rubies is (3/9)(2/8) = 1/12 C it is Intern Joined: 29 Aug 2013 Posts: 12 Re: A certain bag of gemstones is composed of two-thirds [#permalink] ### Show Tags 09 Jan 2014, 15:27 Can someone explain the 2/3 * (2R-1)/(3R-1) part? EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12189 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A certain bag of gemstones is composed of two-thirds [#permalink] ### Show Tags 03 Feb 2015, 22:53 2 1 Hi All, We can solve this problem by TESTing VALUES. However, we have so much specific information, we CANNOT TEST random values. We have to use the information in the prompt to pick a logical number that matches all of the given "restrictions" Here's what we have to work with: 1) Since the gems can be broken down into 2/3 diamonds and 1/3 rubies, the TOTAL must be a MULTIPLE of 3. 2) Since the probability of pulling 2 diamonds is 5/12, when we multiply the two individual probabilities, we MUST end with a denominator that is a multiple of 12 (so the fraction can be reduced to 5/12). Let's start at "3" and work up.... If there are 3 gems, then we have 2 diamonds. The probability of pulling 2 diamonds is (2/3)(1/2) = 2/6 which is NOT a match. If there are 6 gems, then we have 4 diamonds. The probability of pulling 2 diamonds is (4/6)(3/5) = 12/30.....30 cannot reduce to 12. This is NOT a match If there are 9 gems, then we have 6 diamonds. The probability of pulling 2 diamonds is (6/9)(5/8) = 5/12...This IS a MATCH So we have.... Total= 9 Diamonds = 6 Rubies = 3 The question asks for the probability of selecting 2 rubies.... The probability of selecting the first ruby = (3/9) The probability of selecting the second ruby = (2/8) (3/9)(2/8) = 6/72 = 1/12 Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: A certain bag of gemstones is composed of two-thirds  [#permalink]

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05 Jan 2017, 20:27
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The simplest way to solve the problem is to recognize that the total number of gems in the bag must be a multiple of 3, since we have 2/3 diamonds and 1/3 rubies. If we had a total number that was not divisible by 3, we would not be able to divide the stones into thirds. Given this fact, we can test some multiples of 3 to see whether any fit the description in the question.
The smallest number of gems we could have is 6: 4 diamonds and 2 rubies (since we need at least 2 rubies). Is the probability of selecting two of these diamonds equal to 5/12?
4/6 × 3/5 = 12/30 = 2/5. Since this does not equal 5/12, this cannot be the total number of gems.
The next multiple of 3 is 9, which yields 6 diamonds and 3 rubies:
6/9 × 5/8 = 30/72 = 5/12. Since this matches the probability in the question, we know we have 6 diamonds and 3 rubies. Now we can figure out the probability of selecting two rubies:
3/9 × 2/8 = 6/72 = 1/12
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Re: A certain bag of gemstones is composed of two-thirds  [#permalink]

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08 Dec 2017, 11:27
BG 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

We are given that the bag contains two-thirds diamonds and one-third rubies. We are also given that the probability of randomly selecting two diamonds from the bag, without replacement, is 5/12.

Let’s represent the number of rubies in the bag by x. Since the number of diamonds is twice the number of rubies, the number of diamonds in the bag will be 2x.

The probability of choosing two diamonds from the bag, without replacement, can be represented in terms of x by (2x/3x)((2x - 1)/(3x - 1)).

We are also given that this probability is equal to 5/12. Thus:

(2x/3x)((2x - 1)/(3x - 1)) = 5/12

(4x^2 - 2x)/(9x^2 - 3x) = 5/12

Cross multiply to obtain:

48x^2 - 24x = 45x^2 - 15x

3x^2 - 9x = 0

3x(x - 3) = 0

From this equation, we obtain x = 0 or x = 3. Since we know the bag is not empty, x must equal 3, and thus there are 6 diamonds and 3 rubies in the bag. Now the probability of choosing two rubies from the bag, without replacement, can be calculated to be (3/9)(2/8) = (1/3)(1/4) = 1/12.

Alternate Solution:

We can let T = the total number of gems, (2/3)T = diamonds, and (1/3)T = rubies.

Since the probability of randomly selecting two diamonds from the bag, without replacement, is 5/12, we can create the following equation:

[(2/3)T/T] x [((2/3)T - 1)/(T - 1)] = 5/12

(2/3) x ((2/3)T - 1)/(T - 1) = 5/12

((2/3)T - 1)/(T - 1) = 5/8

Cross multiply, we have:

(16/3)T - 8 = 5T - 5

16T - 24 = 15T - 15

T = 9

We know that there are twice as many diamonds as rubies. Thus, there are 6 diamonds and 3 rubies, so the probability of selecting two rubies from the bag is:

3/9 x 2/8 = 1/3 x 1/4 = 1/12

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Re: A certain bag of gemstones is composed of two-thirds &nbs [#permalink] 08 Dec 2017, 11:27
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