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
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
(4x^2 - 2x)/(9x^2 - 3x) = 5/12
Cross multiply to obtain:
48x^2 - 24x = 45x^2 - 15x
3x^2 - 9x = 0
3x(x - 3) = 0
From this equation, we obtain x = 0 or x = 3. Since we know the bag is not empty, x must equal 3, and 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 multiply, 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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