quantskillsgmat wrote:A box contains three pair of blue gloves and two pair of green gloves.Each pair consists of a left hand glove and right hand glove.Each of the glove is seprated from its mate and mixed together.If three gloves are randomly selected ,what is prob that a matched set(left and right glove of same color)will be there.
a)3/10 b)23/60 c)7/12 d)41/60 e)5/6

IMO D


kinds of combinations can be

B B B
G G G
B G G
G B B

I. All three are blue

Total number of combinations= 6C3 = 20
No of combinations in which no two of them are pairs = 2 (either all are left or all are right)
Hence, no of combinations such taht we get a atched set = 20-2 = 18

II. All are green

TOtal number of combinations = 4C3 = 4
No of combinations such that we get a matched set = 4 (there are 2 pairs of green in total, so each possible combination we give us a matching pair)


III. One blue, rest green

Total number of combinations = 6*4C2 = 36
we have two left gloves of green color and two right gloves of green color, number of combinations = 4

Number of combinations such that we get a matching set = 6*4 = 24


IV. one green, rest blue

Total number of combinations = 6C2*4 = 60
for blue,
3 left
3 right
number of combinations = 9

NUmber of combinations for a matching pait = 4*9 = 36



Combining the 4 cases,
Total number of combinations = 20+4+36+60 = 120
Total number of combinatiosn such taht we get a matching pair = 18+4+24+36 = 82

Probability = 82/120 = 41/60

Option D

I dont know if there is a shorter method to solve this...

I suppose that one way to save time would be to recognize that the total number of possible combinations = 10C3 = 10!/(7!)(3!) = 120.

quantskillsgmat wrote:A box contains three pair of blue gloves and two pair of green gloves.Each pair consists of a left hand glove and right hand glove.Each of the glove is seprated from its mate and mixed together.If three gloves are randomly selected ,what is prob that a matched set(left and right glove of same color)will be there.
a)3/10 b)23/60 c)7/12 d)41/60 e)5/6

Another approach:
P(matching set) = 1 - P(no matching set).

There are 3 ways to NOT select a matching set of gloves.

Case 1: 3 non-matching blue
P(1st glove is blue) = 6/10.
P(2nd glove is blue and for the same hand) = 2/9.
P(3rd glove is blue and for the same hand) = 1/8.
Since we want all of these events to happen, we multiply the fractions:
6/10 * 2/9 * 1/8 = 1/60.

Case 2: 1 green, 2 non-matching blue
P(1st glove is green) = 4/10.
P(2nd glove is blue) = 6/9.
P(3rd glove is blue and for the same hand as the 2nd glove) = 2/8.
Since we want all of these events to happen, we multiply the fractions:
4/10 * 6/9 * 2/8 = 4/60.
Since the one green glove could be 1st, 2nd or 3rd, we multiply by 3:
4/60 * 3 = 12/60.

Case 3: 1 blue, 2 non-matching green
P(1st glove is blue) = 6/10.
P(2nd glove is green) = 4/9.
P(3rd glove is green and for the same hand as the 2nd glove) = 1/8.
Since we want all of these events to happen, we multiply the fractions:
6/10 * 4/9 * 1/8 = 1/30.
Since the one blue glove could be 1st, 2nd or 3rd, we multiply by 3:
1/30 * 3 = 3/30 = 6/60.

Since any of the 3 cases above would yield no matching set of gloves, we add the probabilities:
P(no matching set) = 1/60 + 12/60 + 6/60 = 19/60.

Thus, P(matching set) = 1 - 19/60 = 41/60.

The correct answer is D.

@GMATGuruNY

Doesn't the case 1 has a pair of 2 blue gloves. If we are taking 3 blue gloves then definitely we will have 1 pair

nikhilgupta wrote:@GMATGuruNY

Doesn't the case 1 has a pair of 2 blue gloves. If we are taking 3 blue gloves then definitely we will have 1 pair

Here's Case 1:

Case 1: 3 non-matching blue
P(1st glove is blue) = 6/10.
P(2nd glove is blue and for the same hand) = 2/9.
P(3rd glove is blue and for the same hand) = 1/8.
Since we want all of these events to happen, we multiply the fractions:
6/10 * 2/9 * 1/8 = 1/60.

Note the words in red. Each glove selected is FOR THE SAME HAND: all three are for the left hand, or all three are for the right hand. Thus, no matching set -- one left and one right -- is selected.