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
