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805+ (Hard)|   Probability|      
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KarishmaB
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A box contains three pairs of blue gloves and two pairs of 12 Jul 2017, 01:43
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ShashankDave
Bunuel
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

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 is 2 same hand gloves left out of 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.

Dear Bunuel, questions of this type actually give me hell.

I face problems in identifying what is unique or identical and what's not!

Here I at first thought that the socks of the same color and foot were identical, but I don't know if that's true. Then I thought that every pair is unique, which I also don't know. Please help me on how to comprehend questions of this type. How to actually translate and understand what's unique and what's not? VeritasPrepKarishma your inputs are also most welcome.
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The question gives you clues:

"Each pair consists of a left-hand glove and a right-hand glove."

Shows that the gloves in a pair are distinct - left hand and right hand

"what is the probability that a matched set (i.e., a left- and right-hand glove of the same color)"

A matched set needs to have a left and right hand glove of the same color. That is the only requirement. So all three left hand blue gloves are identical and all three right hand blue gloves are identical. Same for the green gloves.
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A box contains three pairs of blue gloves and two pairs of 07 Oct 2017, 04:47
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I solved this problem like this,

case 1: 1 Blue Left, 1 Blue right, any green color glove
Probability = (3c1(1 out of 3 blue left) * 3c1 (1 out of 3 blue right) * 4c1(1 out of 4 green gloves)/(10c3) = (3 * 3 * 4)/10c3 = 36/10c3

case 2: 1 Green Left, 1 Green right, any blue color glove
Probability = (2c1 * 2c1 * 6c1)/10c3 = (2 * 2 * 6)/10c3 = 24/10c3

case 3: 2 Blue Left, 1 Blue right OR 1 Blue Left, 2 Blue right
Probability = (3c2 * 3c1)/10c3 + (3c1 * 3c2)/10c3 = (2 * 3 * 3)/10c3 = 18/10c3

case 4: 2 Green Left 1 Green right OR 1 Green Left 2 Green right
Probability = ( 2c2 * 2c1)/10c3 + (2c1 * 2c2)/10c3 = (2 * 1 * 2)/10c3 = 4/10c3

In all above cases, we get a pair of left-right glove of same color.

So Total probability = sum of all above cases = (36 + 24 + 18 + 4)/10c3 = 41/60
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A box contains three pairs of blue gloves and two pairs of 22 Jan 2019, 05:22
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VeritasKarishma
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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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
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A box contains three pairs of blue gloves and two pairs of 23 Jan 2019, 01:54
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ashwini2k6jha
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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.

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
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The Blue pairs are not distinct. So BLeft of all 3 are same. So are BRight of all 3.
When you do 5C1, you are picking one of the 5 pairs in 5 ways. But picking one Blue pair is the same as picking the other Blue pair.
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A box contains three pairs of blue gloves and two pairs of 24 Jul 2019, 02:38
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VeritasKarishma
12bhang
Hi Karishma,

Could you please tell me how I'm going wrong?

Since there are 3 blue pairs and 2 green pairs, we have a total of 10 gloves.

I considered two cases:
Case 1: we have a blue pair match

This can be done if we have a blue left, a blue right and any other glove. So,
Since we have 3 blue lefts and 3 blue rights and a total of 10 gloves,

3C1*3C1*8C1- 3 ways of selecting a blue left, 3 for a blue right and any one glove of the remaining 8=72 ways.

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.


12bhang

Case 2: we have a green pair match,

SO, Gleft,Gright and any other glove,

2C1*2C1*8C1=32

Summing , we get 104

The total number of ways to select 3 gloves =10C3=120

so probability of getting a match=104/120 = 13/15.

Where am i going wrong?

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
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Responding to a pm:
Quote:

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?
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Yes, 10C1*9C1*8C1 implies that we have 10 distinct objects and we have to pick 3 of them and arrange them in 3 distinct slots: first place, second place and third place.
But we don't have any distinct slots. This is a combinations problem. We need to just select, not arrange in any slots.
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A box contains three pairs of blue gloves and two pairs of 17 Mar 2020, 06:44
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VeritasKarishma
metallicafan

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)


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 \(\frac{3}{10}\)

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Your approach is correct if you know why you did what you did. My only question is this: Why did you multiply by 3!/2! instead of 3! (different arrangements of BBG) since the two Bs are different?

If you understand that when you say 6/10 * 3/9, you are already counting in all arrangements of Bleft and Bright and now all you need to do is arrange G with respect to the 2 Bs (i.e. multiply by 3 for the 3 spots where we can put G), then absolutely, go ahead. There is nothing wrong.
(When you say 6/10, you are counting the possibilities of picking a Bleft or a Bright first and whatever is leftover next, so you have already arranged the different Bs.)[/quote]

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?? :dazed :?
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A box contains three pairs of blue gloves and two pairs of 18 Mar 2020, 09:44
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Kritisood
VeritasKarishma
metallicafan

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)


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 \(\frac{3}{10}\)

Quote:

Your approach is correct if you know why you did what you did. My only question is this: Why did you multiply by 3!/2! instead of 3! (different arrangements of BBG) since the two Bs are different?

If you understand that when you say 6/10 * 3/9, you are already counting in all arrangements of Bleft and Bright and now all you need to do is arrange G with respect to the 2 Bs (i.e. multiply by 3 for the 3 spots where we can put G), then absolutely, go ahead. There is nothing wrong.
(When you say 6/10, you are counting the possibilities of picking a Bleft or a Bright first and whatever is leftover next, so you have already arranged the different Bs.)

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?? :dazed :?
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You have accounted for BBG, but how about BGB and GBB? These will give two more 1/10s.
Also, how about BBB (one pair of Bs and another extra B)
Similarly, GGB and GGG.
Check out the complete solution here: https://gmatclub.com/forum/a-box-contai ... l#p1045908
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A box contains three pairs of blue gloves and two pairs of 24 May 2021, 05:59
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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.
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A box contains three pairs of blue gloves and two pairs of 24 May 2021, 11:38
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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.
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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 (D) 41/60 (E) 5/6

BL BR and (BL / BR / GL / GR): 54
(BL BR) BL = 3c1 x 3c1 x 2c1/2! = 9
(BL BR) BR = 3c1 x 3c1 x 2c1/2! = 9
(BL BR) GL = 3c1 x 3c1 x 2c1 = 18
(BL BR) GR = 3c1 x 3c1 x 2c1 = 18

GL GR and (BL / BR / GL / GR) : 28
(GL GR) BL = 2c1 x 2c1 x 3c1 = 12
(GL GR) BR = 2c1 x 2c1 x 3c1 = 12
(GL GR) GL = 2c1 x 2c1 x 1c1/2! = 2
(GL GR) GR = 2c1 x 2c1 x 1c1/2! = 2

Total = 54 + 28 = 82
Probability = 82/10c3 = 82/120 = 41/60

Hope this helps!
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A box contains three pairs of blue gloves and two pairs of 11 May 2024, 03:16
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We have
  • 3 blue left
  • 3 blue right
  • 2 green left
  • 2 green right­

1) 1 Blue pair + 1 Green: 3 * 3 * 4 = 36

2) All 3 are Blue

a) 2 left + 1 right: 3C2 * 3C1 = 3 * 3 = 9
b) 1 left + 1 right: 3C1 * 3C2 = 3 * 3 = 9

=> Total is 18

3) 1 Green pair + 1 Blue: 2 * 2 * 6 = 24

4) All 3 are Green
=> in any case we will get 1 pair
=> 4C3 = 4


==> Total number of possible cases is 36 + 18 + 24 + 4 = 82

Total number of cases is 10C3 = 120

Probability = \(\frac{82}{120} = \frac{41}{60}\)­
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A box contains three pairs of blue gloves and two pairs of 14 Sep 2024, 04:21
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12bhang
Hi Karishma,

Could you please tell me how I'm going wrong?

Since there are 3 blue pairs and 2 green pairs, we have a total of 10 gloves.

I considered two cases:
Case 1: we have a blue pair match

This can be done if we have a blue left, a blue right and any other glove. So,
Since we have 3 blue lefts and 3 blue rights and a total of 10 gloves,

3C1*3C1*8C1- 3 ways of selecting a blue left, 3 for a blue right and any one glove of the remaining 8=72 ways.




Case 2: we have a green pair match,

SO, Gleft,Gright and any other glove,

2C1*2C1*8C1=32

Summing , we get 104

The total number of ways to select 3 gloves =10C3=120

so probability of getting a match=104/120 = 13/15.

Where am i going wrong?
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Bunuel KarishmaB

What is wrong with this approach ?
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A box contains three pairs of blue gloves and two pairs of 11 Mar 2025, 03:40
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It is luring to go for 3C1 x 3C1 x 8C1 + 2C1 x 2C1 x 8C1 = 104
But many repetitions happen in these 104 outcomes.

For example, in the 1st case, let's assume all 3 are blue.
So, L1, L2, L3, R1, R2, R3 are the total options to choose from.
3C1 x 3C1 = (L1 R1), (L1 R2), (L1 R3), (L2 R1), (L2 R2), (L2 R3), (L3 R1), (L3 R2), (L3 R3) = 9 ways.

Now, say, the 3rd one is a Blue left glove. In the above formula, out of total 8 in 8C1, 2 Blue left, 2 Blue right, and 4 Greens remain. So, when we multiply 3C1 x 3C1 x 2C1 (for the 2 Blue left) we get 18 ways (we are not multiplying by 8C1 because we only want outcomes that have 2 Blue left as the 3rd glove):
(L1 R1 L2), (L1 R1 L3),
(L1 R2 L2), (L1 R2 L3),
(L1 R3 L2), (L1 R3 L3),
(L2 R1 L1), (L2 R1 L3),
(L2 R2 L1), (L2 R2 L3),
(L2 R3 L1), (L2 R3 L3),
(L3 R1 L1), (L3 R1 L2),
(L3 R2 L1), (L3 R2 L2),
(L3 R3 L1), (L3 R3 L2).

Upon looking closely, we can see that 1 and 7 are similar, 2 and 13 are similar, and so on. All are counted twice as ORDER DOESN’T MATTER. Similar double counting is possible when we multiply by 2C1 for 2 Blue right as the 3rd glove.

So, to correct this, we should break up our cases:
Case 1: 2 Blue left, 1 Blue right = 3C2 x 3C1 = 9 [No double counting]
Case 2: 1 Blue left, 2 Blue right = 3C1 x 3C2 = 9 [No double counting]
Case 3: 1 Blue left, 1 Blue right, 1 Green = 3C1 x 3C1 x 4C1 = 36
Case 4: 2 Green left, 1 Green right = 2C2 x 2C1 = 2 [No double counting]
Case 5: 1 Green left, 2 Green right = 2C1 x 2C2 = 2 [No double counting]
Case 6: 1 Green left, 1 Green right, 1 Blue = 2C1 x 2C1 x 6C1 = 24

Total favourable ways = 9 + 9 + 36 + 2 + 2 + 24 = 82
Total possible ways to select 3 gloves from 10 (5 pair) = 10C3 = 120
Probability = 82/120 = 41/60

In combinations, while counting, we should break up our cases so that similar type of items is selected together instead of in multiple separate instances. Otherwise, we need to be cognizant of the double counting and divide the outcome accordingly.
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A box contains three pairs of blue gloves and two pairs of 11 Mar 2025, 15:18
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Hi Karishma,

Could you please tell me how I'm going wrong?

Since there are 3 blue pairs and 2 green pairs, we have a total of 10 gloves.

I considered two cases:
Case 1: we have a blue pair match

This can be done if we have a blue left, a blue right and any other glove. So,
Since we have 3 blue lefts and 3 blue rights and a total of 10 gloves,

3C1*3C1*8C1- 3 ways of selecting a blue left, 3 for a blue right and any one glove of the remaining 8=72 ways.




Case 2: we have a green pair match,

SO, Gleft,Gright and any other glove,

2C1*2C1*8C1=32

Summing , we get 104

The total number of ways to select 3 gloves =10C3=120

so probability of getting a match=104/120 = 13/15.

Where am i going wrong?


Bunuel KarishmaB

What is wrong with this approach ?
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The problem with this approach is that a given glove can be both in the first draw AND in the draw from the group of remaining gloves, leading to double counting.

For example, embedded in the possibilities following this approach is:

3 ways to pick a blue left * 2 ways to pick a blue left from the draw from the remaining 8 gloves = 6 ways.

But the proper approach would be to be to treat this as picking 2 blue gloves from 3 or 3C2 = 3 ways.

So the approach of picking from the remaining gloves doesn't work because a given glove could be both in the first pick as well as in the group of remaining gloves.
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A box contains three pairs of blue gloves and two pairs of 07 May 2025, 23:42
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Bunuel
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
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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why would we not do the same for ggg
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A box contains three pairs of blue gloves and two pairs of 08 May 2025, 00:06
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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
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.­

why would we not do the same for ggg
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Because we cannot select 3 green gloves without selecting a pair.

2 green pairs = 2 left-hand green gloves + 2 right-hand green gloves = 4 green gloves in total.

Let's try and select 3 green gloves from these 4 without selecting a pair of them i.e. right and left together.

Say, 1st green glove is a left green glove; 2nd green glove is again a left green glove. Now, we have run out of left-hand green gloves. The 3rd one we choose has to be one of the remaining 2 right-hand green gloves, so we are forced to include a pair in our selection of 3 green gloves.

Try the same exercise by starting with a right green glove and we will get a similar problem.

So, a case where we pick 3 green gloves without including a pair is impossible.
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A box contains three pairs of blue gloves and two pairs of 24 May 2025, 05:02
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why cant we take the case of GGG here ?
Bunuel
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
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 box contains three pairs of blue gloves and two pairs of 24 May 2025, 05:16
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sriramindurthi
why cant we take the case of GGG here ?
Bunuel
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
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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There can't be 3 same hand Green gloves, since we have only two pairs of Green gloves. We will always get one pair when we select 3 Green gloves from 4.
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A box contains three pairs of blue gloves and two pairs of 24 May 2025, 22:46
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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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Hi KarishmaB / Bunuel, hope you are doing good. I love this explanation, and I understand this very clearly. But I have doubt with what i did. I actually considered like this. Firstly, out of the 3 blue pairs I am taking one and out of the remaining 4 gloves in green I am taking one and I ordered them 2 factorial ways. And then I considered taking one out of the blue pair 3C1 and one out of the remaining 4 blue Indvidual gloves i take one as 4C1 and order them in 2! ways. For the 3rd case, i did take a pair from the 2 green glove pair as 2C1 and out of the remaining individual gloves i take 1 as 2C1 and order them in 2! ways. Then the 4th case as 2C1 from a green pair and out of the 6 remaining blue gloves i take 1 as 6C1 and order them in 2! ways. I am not sure why/where i am wrong. My question is should we take a pair as one count or should we always go by individual ones?? Please brief me on this. Also is it possible to have such questions in the actual gmat focus edition?
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A box contains three pairs of blue gloves and two pairs of 25 May 2025, 21:23
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You can't pick a pair. You are given: Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box.


SwethaReddyL
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.
Hi KarishmaB / Bunuel, hope you are doing good. I love this explanation, and I understand this very clearly. But I have doubt with what i did. I actually considered like this. Firstly, out of the 3 blue pairs I am taking one and out of the remaining 4 gloves in green I am taking one and I ordered them 2 factorial ways. And then I considered taking one out of the blue pair 3C1 and one out of the remaining 4 blue Indvidual gloves i take one as 4C1 and order them in 2! ways. For the 3rd case, i did take a pair from the 2 green glove pair as 2C1 and out of the remaining individual gloves i take 1 as 2C1 and order them in 2! ways. Then the 4th case as 2C1 from a green pair and out of the 6 remaining blue gloves i take 1 as 6C1 and order them in 2! ways. I am not sure why/where i am wrong. My question is should we take a pair as one count or should we always go by individual ones?? Please brief me on this. Also is it possible to have such questions in the actual gmat focus edition?
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A box contains three pairs of blue gloves and two pairs of 17 Aug 2025, 20:06
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Hi Bunuel,
I want to ask How do I know that the question asked for set without replacement? (BBG not equal BGB)
Bunuel
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
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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