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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

by BTGmoderatorLU » Sat Jul 20, 2019 5:25 pm

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Source: Manhattan Prep

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 \(\frac{5}{12}\), what is the probability of selecting two rubies from the bag, without replacement?

A. \(\frac{5}{36}\)

B. \(\frac{5}{24}\)

C. \(\frac{1}{12}\)

D. \(\frac{1}{6}\)

E. \(\frac{1}{4}\)

The OA is C
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A certain bag of gemstones is composed of two-thirds

by Jay@ManhattanReview » Sun Jul 21, 2019 9:53 pm
BTGmoderatorLU wrote:Source: Manhattan Prep

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 \(\frac{5}{12}\), what is the probability of selecting two rubies from the bag, without replacement?

A. \(\frac{5}{36}\)

B. \(\frac{5}{24}\)

C. \(\frac{1}{12}\)

D. \(\frac{1}{6}\)

E. \(\frac{1}{4}\)

The OA is C
We are given that the probability of randomly selecting two diamonds from the bag, without replacement, is \(\frac{5}{12}\). With the help of this information, we can get the number of gemstones.

Say there are 3G gemstones; thus, there are 2G diamonds and G rubies.

Probability of selecting two diamonds, without replacement = (2G)C2 / (3G)C2 = [2G(2G - 1)] / [3G(3G - 1)] = [2(2G - 1)] / [3(3G - 1)] =

= [4G - 2] / 3[2G - 1] = 5/12

=> G = 3; number of gemstones = 3G = 9

Thus, there are 2/3 of 9 = 6 diamonds and 1/3 of 9 = 3 rubies.

Probability of selecting two rubies = (# of ways two rubies can be selected) / (# of ways any two gemstones can be selected)

= 3C2 / 9C2 = (3.2) / (9.8) = 1/12

The correct answer: C

Hope this helps!

-Jay
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Last edited by Jay@ManhattanReview on Tue Jul 23, 2019 7:47 pm, edited 1 time in total.
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by regor60 » Mon Jul 22, 2019 7:46 am
Jay@ManhattanReview wrote:
BTGmoderatorLU wrote:Source: Manhattan Prep

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 \(\frac{5}{12}\), what is the probability of selecting two rubies from the bag, without replacement?

A. \(\frac{5}{36}\)

B. \(\frac{5}{24}\)

C. \(\frac{1}{12}\)

D. \(\frac{1}{6}\)

E. \(\frac{1}{4}\)

The OA is C
The information, "If the probability of randomly selecting two diamonds from the bag, without replacement, is \(\frac{5}{12}\)," is redundant.

Let's come to the solution:

Say, there are 9 gemstones; thus, there are 2/3 of 9 = 6 diamonds and 1/3 of 9 = 3 rubies.

Probability of selecting two rubies = (# of ways two rubies can be selected) / (# of ways any two gemstones can be selected)

= 3C2 / 9C2 = (3.2) / (9.8) = 1/12

The correct answer: C


Actually, I believe 5/12 is intrinsic to determining that there are 9 gems, if it weren't one could just as easily suppose 12 gems and follow the logic above and get the wrong answer. Of course, the above works as an example of trial and error getting it right on the first try.

Number of Diamonds = D, total number of gems = N. Probability of selecting two diamonds is therefore (D/N)*((D-1)/(N-1)) = 5/12

so D*(D-1)=5/12N*(N-1) or 12D^2 - 12D = 5N^2-5N

substituting D=(2/3)N and so forth yields N^2-9N = 0 = N(N-9) = 0 or N=9

so 6 diamonds and 3 rubies etc
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by Ian Stewart » Tue Jul 23, 2019 7:24 am
Jay@ManhattanReview wrote: The information, "If the probability of randomly selecting two diamonds from the bag, without replacement, is \(\frac{5}{12}\)," is redundant.
Yes, regor60 is correct - this information is not redundant. You can see that the total number of gems influences the answer just by imagining having 3 trillion gems in total. Then the probability of picking two diamonds will be negligibly less than (2/3)(2/3) = 4/9, so will not equal 5/12. We need to use the 5/12 to work out the total number of gems.

There is, however, a fundamental problem with the sentence "If the probability of randomly selecting two diamonds from the bag, without replacement, is \(\frac{5}{12}\)". The natural question to ask, reading that sentence, is when doing what?. You can't describe random selections in this way.
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by Jay@ManhattanReview » Tue Jul 23, 2019 7:39 pm
regor60 wrote:
Jay@ManhattanReview wrote:
BTGmoderatorLU wrote:Source: Manhattan Prep

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 \(\frac{5}{12}\), what is the probability of selecting two rubies from the bag, without replacement?

A. \(\frac{5}{36}\)

B. \(\frac{5}{24}\)

C. \(\frac{1}{12}\)

D. \(\frac{1}{6}\)

E. \(\frac{1}{4}\)

The OA is C
The information, "If the probability of randomly selecting two diamonds from the bag, without replacement, is \(\frac{5}{12}\)," is redundant.

Let's come to the solution:

Say, there are 9 gemstones; thus, there are 2/3 of 9 = 6 diamonds and 1/3 of 9 = 3 rubies.

Probability of selecting two rubies = (# of ways two rubies can be selected) / (# of ways any two gemstones can be selected)

= 3C2 / 9C2 = (3.2) / (9.8) = 1/12

The correct answer: C


Actually, I believe 5/12 is intrinsic to determining that there are 9 gems, if it weren't one could just as easily suppose 12 gems and follow the logic above and get the wrong answer. Of course, the above works as an example of trial and error getting it right on the first try.

Number of Diamonds = D, total number of gems = N. Probability of selecting two diamonds is therefore (D/N)*((D-1)/(N-1)) = 5/12

so D*(D-1)=5/12N*(N-1) or 12D^2 - 12D = 5N^2-5N

substituting D=(2/3)N and so forth yields N^2-9N = 0 = N(N-9) = 0 or N=9

so 6 diamonds and 3 rubies etc
Yes, regor60. You are correct. The information is need.
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by Scott@TargetTestPrep » Tue Jul 30, 2019 8:36 am
BTGmoderatorLU wrote:Source: Manhattan Prep

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 \(\frac{5}{12}\), what is the probability of selecting two rubies from the bag, without replacement?

A. \(\frac{5}{36}\)

B. \(\frac{5}{24}\)

C. \(\frac{1}{12}\)

D. \(\frac{1}{6}\)

E. \(\frac{1}{4}\)

The OA is C

We are given that the bag contains two-thirds diamonds and one-third rubies. We are also given that the probability of randomly selecting two diamonds from the bag, without replacement, is 5/12.

Let's represent the number of rubies in the bag by x. Since the number of diamonds is twice the number of rubies, the number of diamonds in the bag will be 2x.

The probability of choosing two diamonds from the bag, without replacement, can be represented in terms of x by (2x/3x)((2x - 1)/(3x - 1)).

We are also given that this probability is equal to 5/12. Thus:

(2x/3x)((2x - 1)/(3x - 1)) = 5/12

Since we know the bag is not empty, x must be non-zero; therefore, we can cancel the x's in the first fraction:

(2/3)((2x - 1)/(3x - 1)) = 5/12

(4x - 2)/(9x - 3) = 5/12

Cross multiply to obtain:

48x - 24 = 45x - 15

3x = 9

x = 3

Thus, there are 6 diamonds and 3 rubies in the bag. Now, the probability of choosing two rubies from the bag, without replacement, can be calculated to be (3/9)(2/8) = (1/3)(1/4) = 1/12.

Alternate Solution:

We can let T = the total number of gems, (2/3)T = diamonds, and (1/3)T = rubies.

Since the probability of randomly selecting two diamonds from the bag, without replacement, is 5/12, we can create the following equation:

[(2/3)T/T] x [((2/3)T - 1)/(T - 1)] = 5/12

(2/3) x ((2/3)T - 1)/(T - 1) = 5/12

((2/3)T - 1)/(T - 1) = 5/8

Cross multiplying, we have:

(16/3)T - 8 = 5T - 5

16T - 24 = 15T - 15

T = 9

We know that there are twice as many diamonds as rubies. Thus, there are 6 diamonds and 3 rubies, so the probability of selecting two rubies from the bag is:

3/9 x 2/8 = 1/3 x 1/4 = 1/12

Answer: C

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