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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, 07:32
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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
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Re: Challenge MHTNGMAT [#permalink] New post 29 May 2006, 09:03
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, 09:46
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, 17:38
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, 19:23, 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, 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 :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, 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 :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] 09 Feb 2013, 19:28
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