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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...

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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

The OA is C.

I'm really confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.

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DavidG@VeritasPrep wrote:
LUANDATO 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

The OA is C.

I'm really confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.
We're told that there are twice as many diamonds as rubies in this bag, so let's designate the number of diamonds as '2x' and the number of rubies as 'x,' giving us a total of 3x.

Probability that the first pick is a diamond = # diamonds/# tot gems = 2x/3x
Probability that the second pick is also a diamond given that the first pick was a diamond = (2x -1)/(3x -1)
Probability that the two picks are diamonds = (2x/3x) * (2x -1)/(3x -1) = 5/12
(2/3) * (2x -1)/(3x -1) = 5/12
(4x - 2)/(9x -3) = 5/12
48x -24 = 45x - 15
3x = 9
x = 3

So we've got 2x = 2*3 = 6 diamonds, x = 3 rubies and 9 total gems.

We want the probability of selecting two rubies.
Probability that first pick is a ruby = 3/9
Probability that second pick is a ruby given that first pick was a ruby = 2/8
Probability that both picks are rubies = (3/9)(2/8) = 6/72 = 1/12. The answer is C
Note that you could also skip the algebra and just play with scenarios. If we know that there are twice as many diamonds as rubies, we can try:
Scenario One: 2 diamonds and 1 ruby. P(selecting two diamonds) = (2/3)(1/2) = 2/6 --> not 5/12
Scenario Two: 4 diamonds and 2 rubies. P(selecting two diamonds) = (4/6)(3/5) = 12/30 --> not 5/12
Scenario Three: 6 diamonds and 3 rubies. P(selecting two diamonds) = (6/9)(5/8) = 30/72 = 5/12. Perfect! So we know there are 6 diamonds and 3 rubies.

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LUANDATO 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

The OA is C.

I'm really confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.
We're told that there are twice as many diamonds as rubies in this bag, so let's designate the number of diamonds as '2x' and the number of rubies as 'x,' giving us a total of 3x.

Probability that the first pick is a diamond = # diamonds/# tot gems = 2x/3x
Probability that the second pick is also a diamond given that the first pick was a diamond = (2x -1)/(3x -1)
Probability that the two picks are diamonds = (2x/3x) * (2x -1)/(3x -1) = 5/12
(2/3) * (2x -1)/(3x -1) = 5/12
(4x - 2)/(9x -3) = 5/12
48x -24 = 45x - 15
3x = 9
x = 3

So we've got 2x = 2*3 = 6 diamonds, x = 3 rubies and 9 total gems.

We want the probability of selecting two rubies.
Probability that first pick is a ruby = 3/9
Probability that second pick is a ruby given that first pick was a ruby = 2/8
Probability that both picks are rubies = (3/9)(2/8) = 6/72 = 1/12. The answer is C

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