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A certain bag of gemstones is composed of two-thirds [#permalink]
29 May 2006, 06:32

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Difficulty:

75% (hard)

Question Stats:

59% (03:06) correct
41% (02:16) wrong based on 187 sessions

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?

Re: Challenge MHTNGMAT [#permalink]
29 May 2006, 08:03

1

This post received 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

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: Challenge MHTNGMAT [#permalink]
29 May 2006, 08:46

2

This post received KUDOS

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

Re: A certain bag of gemstones is composed of two-thirds [#permalink]
06 Feb 2013, 16:38

1

This post received 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

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 _________________

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

Last edited by Rock750 on 09 Feb 2013, 18:23, edited 1 time in total.

Re: A certain bag of gemstones is composed of two-thirds [#permalink]
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 though you got it right later

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

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

Re: A certain bag of gemstones is composed of two-thirds [#permalink]
09 Jan 2014, 14:00

2

This post received 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?

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

1

This post received KUDOS

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 _________________

Re: A certain bag of gemstones is composed of two-thirds [#permalink]
03 Feb 2015, 08:56

Hello from the GMAT Club BumpBot!

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

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...