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Probability... again

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Probability... again

by Halimah_O » Mon Dec 05, 2016 6:35 am
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]
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by regor60 » Mon Dec 05, 2016 8:24 am
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]
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.

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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by DavidG@VeritasPrep » Mon Dec 05, 2016 8:39 am
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]
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.

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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by DavidG@VeritasPrep » Mon Dec 05, 2016 8:43 am
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]
Algebraically

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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by Matt@VeritasPrep » Thu Dec 08, 2016 8:22 pm
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.
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by Matt@VeritasPrep » Thu Dec 08, 2016 8:25 pm
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.
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by Anaira Mitch » Thu Jan 05, 2017 7:27 pm
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.
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by [email protected] » Fri Jan 06, 2017 10:30 am
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

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