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A certain bag of gemstones is composed of two-thirds

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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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Question Stats:

63% (03:05) correct 37% (02:26) wrong based on 300 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
[Reveal] Spoiler: OA
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29 May 2006, 09:03
2
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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

2/3 * (2X-1) / (3X-1) = 5/ 12 => X = 3
So total gems = 9
and probability of ruby = 1/3 * 2/8 = 1/12
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29 May 2006, 09:46
2
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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

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

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06 Feb 2013, 17:38
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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 :

$$\frac{1}{3}*\frac{2}{8}=\frac{1}{12}$$

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Last edited by Rock750 on 09 Feb 2013, 19:23, edited 1 time in total.
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Re: A certain bag of gemstones is composed of two-thirds [#permalink]

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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 :

$$\frac{1}{3}*\frac{2}{8}=\frac{1}{12}$$

Looks like you got your diamonds and rubies mixed up though you got it right later
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Re: A certain bag of gemstones is composed of two-thirds [#permalink]

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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 :

$$\frac{1}{3}*\frac{2}{8}=\frac{1}{12}$$

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

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09 Jan 2014, 15:00
3
KUDOS
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
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Re: A certain bag of gemstones is composed of two-thirds [#permalink]

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09 Jan 2014, 15:27
Can someone explain the

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

part?
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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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Expert's post
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews GMAT Club Legend Joined: 09 Sep 2013 Posts: 12197 Followers: 541 Kudos [?]: 151 [0], given: 0 Re: A certain bag of gemstones is composed of two-thirds [#permalink] Show Tags 03 Feb 2015, 09:56 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 7711 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 343 Kudos [?]: 2282 [1] , given: 162 Re: A certain bag of gemstones is composed of two-thirds [#permalink] Show Tags 03 Feb 2015, 22:53 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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: [Reveal] Spoiler: C GMAT assassins aren't born, they're made, Rich _________________ Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests

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

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28 Feb 2016, 10:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: A certain bag of gemstones is composed of two-thirds   [#permalink] 28 Feb 2016, 10:35
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