**Question:**

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left-hand glove and right-hand glove of the same color) will be among the three gloves selected?

A. 3/10

B. 23/60

C. 7/12

D. 41/60

E. 5/6

**MBA Wisdom's Answer:**

$$\text{Gloves:}$$ $$\text{LB = left blue}$$ $$\text{RB = right blue}$$ $$\text{LG = left green}$$ $$\text{RG = right green}$$ $$$$ $$\text{No. of Gloves:}$$ $$\text{LB: 3}$$ $$\text{RB: 3}$$ $$\text{LG: 2}$$ $$\text{RG: 2}$$ $$$$ $$\text{Probability(Match)} = {1} - \text{Probability(No Match)}$$ $$$$ $$\text{No match situations:}$$ $$\text{Probability(LB, LB, LB)} = \left( \frac{3}{10} \right) \left( \frac{2}{9} \right) \left( \frac{1}{8} \right) = \frac{1}{120}$$ $$\text{Probability(RB, RB, RB)} = \frac{1}{120} \textit{ (same as above)}$$ $$\text{Probability(LB, LB, and a G)}= \left( {3} \right) \left( \frac{3}{10} \right) \left( \frac{2}{9} \right) \left( \frac{4}{8} \right) = \frac{1}{10}$$ $$\text{Probability(RB, RB, and a G)} = \frac{1}{10} \textit{ (same as above)}$$ $$\text{Probability(LG, LG, and a B)} = \left( {3} \right) \left( \frac{2}{10} \right) \left( \frac{1}{9} \right) \left( \frac{6}{8} \right) = \frac{1}{20}$$ $$\text{Probability(RG, RG, and a B)} = \frac{1}{20} \textit{ (same as above)}$$

$$\text{Probability(No Match)} = \left(\frac{1}{120} + \frac{1}{120} + \frac{1}{10} + \frac{1}{10} + \frac{1}{20} + \frac{1}{20} \right) = \left(\frac{1}{60} + \frac{12}{60} + \frac{6}{60} \right) = \frac{19}{60}$$

$$\text{Probability(Match)} = {1} - \frac{19}{60} = \frac{41}{60}$$

**D. 41/60**