A box contains three pairs of blue gloves and two pairs of

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

You can use the 'calculating the reverse' method used by Bunuel above or if you would like to calculate the probability of getting a matched set in the usual way, you can think of it in this way:

Bleft (3), Bright(3), Gleft (2), Gright(2)

Bleft, Bright, G
Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways.
Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3!
(You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Bleft, Bright, B
Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any blue in 4/8 ways.
Probability of getting B pair and another B = (3/10)*(3/9)*(4/8)*3!/2!
(You multiply by 3! here to account for the order e.g. Bright, Bleft, Bleft or Bleft, Bright, Bright etc but two gloves will be identical so you divide by 2!)

Gleft, Gright, B
Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any Blue in 6/8 ways.
Probability of getting G pair and a B = (2/10)*(2/9)*(6/8)*3!

Gleft, Gright, G
Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any other G in 2/8 ways.
Probability of getting G pair and a G = (2/10)*(2/9)*(2/8)*3!/2!

Adding them all up, you get 41/60.

Note here that we cannot say that let's get Bleft, Bright and then any one of the remaining gloves. We need to take separate cases for the third glove (B or G i.e. first two cases above) because the number of arrangements of Bleft, Bright, G is different from number of arrangements of Bleft, Bright, B as we see above. In one case we multiply by 3! because all 3 gloves are distinct. In the other case, we multiply by 3!/2! because 2 of the gloves are identical. Same logic can be used for the green pair.

Please correct me why it's wrong to solve in this way..

Select one pair among 5 pair and select one from balance 8 gloves.

5C1*8C1/10C3

thanks in advance

You are double counting here. Say the gloves are all distinct. The 3 blue left ones are Bl1, Bl2 and Bl3. Three blue right ones are Br1, Br2, Br3.
So you select one of the blue left ones and one of the blue right ones: Bl2, Br3.
Now you have 8 leftover and you can select any one of them. Say you select Bl1.
So your selection consists of Bl1, Bl2, Br3

Imagine another scenario:
So you select one of the blue left ones and one of the blue right ones: Bl1, Br3.
Now you have 8 leftover and you can select any one of them. Say you select Bl2.
So your selection consists of Bl1, Bl2, Br3

The two selections are the same but you have counted them as different selections.




Same problem with the green pair.
From the solutions given above, review how to effectively use probability to solve this question.

In case you want to use combinations, you still have to take cases:

All three Blues:
3C2*3C1*2 = 18(Select 2 of the blue left and one of the blue right. Multiply by 2 because you can select 2 of the blue right and one of blue left too)

2 Blues, 1 Green:
3C1*3C1*4C1 = 36

2 Greens, 1 Blue
2C1*2C1*6C1 = 24

Three Greens
2C2*2C1*2 = 4

Total = 82

Select 3 gloves from 10 in 10C3 ways = 120

Probability = 82/120 = 41/60

One quick question on below explanation - When we are calculating total number of case to pick 3 gloves why aren't we doing 10C1*9C1*8C1 (case 1) instead of 10C3 (case 2). I do understand that case 1 is arrangement and case 2 is pure selection but when we calculate the number of case for say 2G and 1B aren't we doing arrangement somehow?

First we are going to see the probabilities when we have a pair of blue gloves =

P(right or left blue glove)* P(the opposite hand blue glove)*P(a green glove) =

\(\frac{6}{10} * \frac{3}{9} * \frac{4}{8} = \frac{1}{10}\)

But because there could be different arragements of BBG, we multiply 1/10 by 3!/2!
So, we have \(\frac{3}{10}\)

Hi VeritasKarishma, I followed the above method. For getting BBG our prob is 1/10 for GGB it will be 4/10*2/9*6/8 = 2/30

adding both we get 1/10+2/30 = 5/30

why is this wrong??

Hmmm, I got very close here. Wonder what I missed?
B = Blue, G = Green, L = Left, R = Right
Total cases = 10C3 = 120
BL + BR can be combined with GR + GL in (2 + 2)*6 ways and with BL + BR in (2 + 2)*3 ways = 36 ways
GL + GR can be combined with BL + BR in (3 + 3)*6 ways and with GR + GL in (1 + 1)*3 ways = 42 ways
In total 78/120 or 39/60
Somewhere I miss 4 cases.

You can use the 'calculating the reverse' method used by Bunuel above or if you would like to calculate the probability of getting a matched set in the usual way, you can think of it in this way:

Bleft (3), Bright(3), Gleft (2), Gright(2)

Bleft, Bright, G
Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any green in 4/8 ways.
Probability of getting B pair and a G = (3/10)*(3/9)*(4/8)*3!
(You multiply by 3! here because you could pick in some other order e.g. Bright, Bleft, G or Bleft, G, Bright etc)

Bleft, Bright, B
Get a Bleft and Bright in (3/10)*(3/9) ways. Then get any blue in 4/8 ways.
Probability of getting B pair and another B = (3/10)*(3/9)*(4/8)*3!/2!
(You multiply by 3! here to account for the order e.g. Bright, Bleft, Bleft or Bleft, Bright, Bright etc but two gloves will be identical so you divide by 2!)

Gleft, Gright, B
Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any Blue in 6/8 ways.
Probability of getting G pair and a B = (2/10)*(2/9)*(6/8)*3!

Gleft, Gright, G
Get a Gleft and Gright in (2/10)*(2/9) ways. Then get any other G in 2/8 ways.
Probability of getting G pair and a G = (2/10)*(2/9)*(2/8)*3!/2!

Adding them all up, you get 41/60.

Note here that we cannot say that let's get Bleft, Bright and then any one of the remaining gloves. We need to take separate cases for the third glove (B or G i.e. first two cases above) because the number of arrangements of Bleft, Bright, G is different from number of arrangements of Bleft, Bright, B as we see above. In one case we multiply by 3! because all 3 gloves are distinct. In the other case, we multiply by 3!/2! because 2 of the gloves are identical. Same logic can be used for the green pair.