arps wrote:

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- 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

Gloves:

LB = left blue

RB = right blue

LG = left green

RG = right green

LB: 3

RB: 3

LG: 2

RG: 2

Approach: 1 - P(no match)

No match situations:

P(LB, LB, LB): (3/10)(2/9)(1/8) = 1/120

P(RB, RB, RB): 1/120

P(LB, LB and a G): (3/10)(2/9)(4/8)*3 (for the 3 different ways we can pull the gloves) = 1/10

P(RB, RB and a G): 1/10 (same as above)

P(LG, LG, and a B): (2/10)(1/9)(6/8)*3 (for the 3 different ways we can pull the gloves) = 1/20

P(RG, RG, and a B): 1/20 (same as above)

P(no match) = 2(1/120) + 2(1/10) + 2(1/20) = 1/60 + 12/60 + 6/60 = 19/60

P(match) = 1 - P(no match) = 1 - 19/60 = 41/60

_________________

Dan Morgan

MBA Wisdom

http://www.mbawisdom.com