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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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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 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

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

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 :

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

Answer : C
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Last edited by Rock750 on 09 Feb 2013, 19:23, edited 1 time in total.

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 :

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

Answer : C

Looks like you got your diamonds and rubies mixed up though you got it right later

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 :

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

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
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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

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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?

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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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).

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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

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).

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