jagdeep wrote: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- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10
(B) 23/60
(C) 7/12
[spoiler](D) 41/60[/spoiler]
(E) 5/6

Found this great explanation on web. Please see the attachment.

jagdeep wrote: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- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10
(B) 23/60
(C) 7/12
[spoiler](D) 41/60[/spoiler]
(E) 5/6

Let me call B for remaining blue right-hand, B' for remaining blue left-hand, B* for remaining blue any hand, G for remaining green right-hand, G' for remaining green left-hand, and G* for remaining green any hand.

Now, BB'G* can be permuted in 3*3*4*3! or 36*6 ways; BB'B and BB'B' each can be permuted in 3*3*2*(3!/2!) ways or a total of 18*6 ways for these selections to have; GG'B* can be permuted in 2*2*6*3! or 24*6 ways; GG'G and GG'G' each can be permuted in 2*2*1*(3!/2!) ways or a total of 4*6 ways for these selections to have.

Above discussed selections are those that confirm that any pair and an odd glove has been obtained. From 10 gloves, 3 gloves can be had in 10*9*8 ways, out of which, 36*6 +18*6 + 24*6 + 4*6 = 82*6 ways favor the chance of any pair and an odd glove; hence the required probability = (82*6)/(10*9*8) = [spoiler]41/60. D[/spoiler]

Hey Sanju..

am a bit confused with your explanation.
Can you be mroe clear..

Thanks Sanju your approach is very good..
Basic idea is the same as Sanju's but used combinations for finding "no pair" draws.

B -> Blue right glove (3)
B' ->Blue left glove (3)
B*-> Blue gloves can be left or right
R -> Red right glove (2)
R' -> Red left glove (2)
R*->Red gloves can be left or right

No. of ways 3 gloves can be drawn from 10 = 10C3 = 120

Ways three gloves can be drawn and not find a pair

1) BBG* = 3C2 * 4C1 = 12
2) B'B'G* = 3C2 * 4C1 = 12
3) BBB = 3C3 =1
4) B'B'B' = 3C3=1
5) GGB* = 2C2 *6C1 = 6
6) G'G'B* = 2C2 * 6C1 =6
Since there are only 2 pairs of green gloves GGG or G'G'G' cannot be a combination.

Adding 1 to 6, no. of ways of not drawing a pair = 38

No. of ways of drawing a pair = 120 - 38 = 82.

Required probability = 82/120 = 41/60

is there any other way to do this problem? as I don't understand the logic that is showcased here.