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
IMO B[/spoiler]
Probability... again
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The probability of selecting the first diamond is 2/3. . What you don't know is the probability of selecting the second diamond, but you're told that the probability of selecting two diamonds is 5/12.Halimah_O 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
IMO B[/spoiler]
5/12 = 2/3 multiplied by X ( probability of selecting second diamond)
X = 5/8. Now you know that because of no replacement, there could have been originally 6 diamonds (5+1) and 9 (8+1) gemstones, so 3 rubies.
With 3 rubies, probability of selecting the first one is 3/9. Probability of selecting second one is 2/8.
Probability of selecting two rubies is therefore 3/9 x 2/8 = 1/12
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You could test cases until we find one that satisfies the conditions of the problem. We know that if the bag is 2/3 diamonds and 1/3 rubies that there are twice as many diamonds as rubies.Halimah_O 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
IMO B[/spoiler]
Case 1: 4 Diamonds and 2 Rubies
P(selecting 2 diamonds) = (4/6) * (3/5) = 12/30 = 2/5. We want 5/12. Close
Case 2: 6 Diamonds and 3 Rubies
P(selecting 2 diamonds) = (6/9) * (5/8) = 30/72 = 5/12. That's what we want.
If there are 3 rubies and 6 diamonds, P(selecting 2 rubies) = (3/9) * (2/8) = 6/72 = 1/12. Answer is C (Please edit the original post.)
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AlgebraicallyHalimah_O 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
IMO B[/spoiler]
Diamonds: 2x
Rubies: x
Total: 3x
P(selecting 2 diamonds) = (2x/3x) * (2x-1)/(3x-1) = 5/12 --> (The x's in 2x/3x will cancel out, then we can cross multiply) --> x = 3
If x = 3,
Diamonds: 2x = 2*3 = 6
Rubies: x = 3
Total: 3x = 3*3 = 9
If there are 3 rubies and 6 diamonds, P(selecting 2 rubies) = (3/9) * (2/8) = 6/72 = 1/12. Answer is C
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If we know that D * (D - 1) / ((D + R) * (D + R - 1)) = 5/12, we could look for friendly values of D that give us this result.
Start by looking for two consecutive integers whose product is a multiple of 12. (This is what we need for the denominator.) 3 * 4 works, but then the numerators won't work out. (We can't get integer * integer = 5 if both of those integers are less than 4!)
The next possibility is 8 * 9. If our answer is x/72 = 5/12, then x = 30. We need D * (D - 1) = 30, so D = 6, and then, voila, we're set!
D = 6, and (D + R) = 9, so there are six diamonds and three rubies.
From there, the probability of selecting two rubies = (3 / 9) * (2 / 8) = 1/12.
Start by looking for two consecutive integers whose product is a multiple of 12. (This is what we need for the denominator.) 3 * 4 works, but then the numerators won't work out. (We can't get integer * integer = 5 if both of those integers are less than 4!)
The next possibility is 8 * 9. If our answer is x/72 = 5/12, then x = 30. We need D * (D - 1) = 30, so D = 6, and then, voila, we're set!
D = 6, and (D + R) = 9, so there are six diamonds and three rubies.
From there, the probability of selecting two rubies = (3 / 9) * (2 / 8) = 1/12.
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Another approach:
We know there are half as many rubies as diamonds, so R = D/2.
From there, we're told that
D/(D + R) * (D - 1)/(D + R - 1) = 5/12.
Substituting D for 2R, we get
2R/(2R + R) * (2R - 1)/(2R + R - 1) = 5/12
or
2/3 * (2R - 1)/(3R - 1) = 5/12
or
(2R - 1)/(3R - 1) = 15/24 = 5/8
and
R = 3
From there, we know R = 3, D = 6, and the computation is easy.
We know there are half as many rubies as diamonds, so R = D/2.
From there, we're told that
D/(D + R) * (D - 1)/(D + R - 1) = 5/12.
Substituting D for 2R, we get
2R/(2R + R) * (2R - 1)/(2R + R - 1) = 5/12
or
2/3 * (2R - 1)/(3R - 1) = 5/12
or
(2R - 1)/(3R - 1) = 15/24 = 5/8
and
R = 3
From there, we know R = 3, D = 6, and the computation is easy.
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The simplest way to solve the problem is to recognize that the total number of gems in the bag must be a multiple of 3, since we have 2/3 diamonds and 1/3 rubies. If we had a total number that was not divisible by 3, we would not be able to divide the stones into thirds. Given this fact, we can test some multiples of 3 to see whether any fit the description in the question.
The smallest number of gems we could have is 6: 4 diamonds and 2 rubies (since we need at least 2 rubies). Is the probability of selecting two of these diamonds equal to 5/12?
4/6 × 3/5 = 12/30 = 2/5. Since this does not equal 5/12, this cannot be the total number of gems.
The next multiple of 3 is 9, which yields 6 diamonds and 3 rubies:
6/9 × 5/8 = 30/72 = 5/12. Since this matches the probability in the question, we know we have 6 diamonds and 3 rubies. Now we can figure out the probability of selecting two rubies:
3/9 × 2/8 = 6/72 = 1/12
The correct answer is C.
The smallest number of gems we could have is 6: 4 diamonds and 2 rubies (since we need at least 2 rubies). Is the probability of selecting two of these diamonds equal to 5/12?
4/6 × 3/5 = 12/30 = 2/5. Since this does not equal 5/12, this cannot be the total number of gems.
The next multiple of 3 is 9, which yields 6 diamonds and 3 rubies:
6/9 × 5/8 = 30/72 = 5/12. Since this matches the probability in the question, we know we have 6 diamonds and 3 rubies. Now we can figure out the probability of selecting two rubies:
3/9 × 2/8 = 6/72 = 1/12
The correct answer is C.
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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: C
GMAT assassins aren't born, they're made,
Rich
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: C
GMAT assassins aren't born, they're made,
Rich