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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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New post 29 May 2006, 06:32
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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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New post 29 May 2006, 08:03
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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


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
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New post 29 May 2006, 08:46
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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


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

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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 :

\(\frac{1}{3}*\frac{2}{8}=\frac{1}{12}\)

Answer : C
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Last edited by Rock750 on 09 Feb 2013, 18:23, 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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New post 09 Feb 2013, 18:13
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 :wink: 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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New post 09 Feb 2013, 18:28
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 :wink: 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 :-D
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Re: A certain bag of gemstones is composed of two-thirds [#permalink]

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New post 09 Jan 2014, 14:00
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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


(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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New post 09 Jan 2014, 14:27
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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New post 09 Jan 2014, 20:55
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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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Re: A certain bag of gemstones is composed of two-thirds [#permalink]

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

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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


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

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

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New post 05 Jan 2017, 19:27
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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
The correct answer is C.
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Re: A certain bag of gemstones is composed of two-thirds   [#permalink] 05 Jan 2017, 19:27
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