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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] 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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Re: Challenge MHTNGMAT [#permalink] 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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Re: Challenge MHTNGMAT [#permalink] 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] New post 06 Feb 2013, 16:38
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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 :

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

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] 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] 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] 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] 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] 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] New post 03 Feb 2015, 08:56
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Re: A certain bag of gemstones is composed of two-thirds [#permalink] New post 03 Feb 2015, 21:53
Expert's post
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] 03 Feb 2015, 21:53
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