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805+ Level|   Probability|      
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Bunuel
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A box contains three pairs of blue gloves and two pairs of 18 Aug 2025, 00:17
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DangPham9625
Hi Bunuel,
I want to ask How do I know that the question asked for set without replacement? (BBG not equal BGB)
Bunuel
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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
Let's calculate the opposite probability of NOT getting a matched set and subtract this value from 1.

This could happen only if we pick all three same hand BLUE gloves; two same hand BLUE gloves and any green glove; or two same hand GREEN gloves and any BLUE glove

BBB: \(\frac{6}{10}*\frac{2}{9}*\frac{1}{8}=\frac{1}{60}\) (after we pick a blue glove, 6/10, then there are 2 same hand gloves left out of the total 9 gloves - 2/9, and so on);

BBG: \((\frac{6}{10}*\frac{2}{9}*\frac{4}{8})*3=\frac{12}{60}\), multiplying by 3 as this scenario can occur in 3 different ways: BBG, BGB, GBB;

GGB: \((\frac{4}{10}*\frac{1}{9}*\frac{6}{8})*3=\frac{6}{60}\);

\(P=1-(\frac{1}{60}+\frac{12}{60}+\frac{6}{60})=\frac{41}{60}\).

Answer: D.

Hope it's clear.­
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A proper GMAT question would always state explicitly whether the drawing is with or without replacement. In this question, it is not stated, but from the context it’s clear that it means the without replacement case.

Also, this has nothing to do with the BBG vs. BGB issue. Those cases are treated separately simply because they are different draw orders. Even if the problem were with replacement, BBG and BGB would still count as different outcomes.
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veritatisvoluptas
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A box contains three pairs of blue gloves and two pairs of 18 Aug 2025, 09:02
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You multiplied Bleft Bright G by 3! In the first step. Why are we doing so, wouldn’t this mean that order does matter or permutation whereas in the question it’s “selected” which means just combination.
BL BR G is just a way, why do we care of how it’s arranged
KarishmaB
arps
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.
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A box contains three pairs of blue gloves and two pairs of 20 Oct 2025, 02:07
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why does arrangement matter in this question?
KarishmaB


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.
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