Try this if it is more helpful
The straight way you want to achieve. NOT always a question must be attacked in ONE way. You must use the most EFFICIENT way
Quote:
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.
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