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Re: A box contains three pairs of blue gloves and two pairs of
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11 Jul 2017, 01:23
Bunuel wrote: arps wrote: A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a lefthand glove and a righthand 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 righthand 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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Re: A box contains three pairs of blue gloves and two pairs of
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12 Jul 2017, 01:43
ShashankDave wrote: Bunuel wrote: arps wrote: A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a lefthand glove and a righthand 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 righthand 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. The question gives you clues: "Each pair consists of a lefthand glove and a righthand 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 righthand 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
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07 Oct 2017, 04:47
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 leftright glove of same color.
So Total probability = sum of all above cases = (36 + 24 + 18 + 4)/10c3 = 41/60



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Re: A box contains three pairs of blue gloves and two pairs of
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08 Apr 2018, 19:49
arps wrote: A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a lefthand glove and a righthand 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 righthand 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 Gloves: LB = left blue RB = right blue LG = left green RG = right green LB: 3 RB: 3 LG: 2 RG: 2 Approach: 1  P(no match) No match situations: P(LB, LB, LB): (3/10)(2/9)(1/8) = 1/120 P(RB, RB, RB): 1/120 P(LB, LB and a G): (3/10)(2/9)(4/8)*3 (for the 3 different ways we can pull the gloves) = 1/10 P(RB, RB and a G): 1/10 (same as above) P(LG, LG, and a B): (2/10)(1/9)(6/8)*3 (for the 3 different ways we can pull the gloves) = 1/20 P(RG, RG, and a B): 1/20 (same as above) P(no match) = 2(1/120) + 2(1/10) + 2(1/20) = 1/60 + 12/60 + 6/60 = 19/60 P(match) = 1  P(no match) = 1  19/60 = 41/60
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Re: A box contains three pairs of blue gloves and two pairs of
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22 Jan 2019, 05:22
VeritasKarishma wrote: arps wrote: A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a lefthand glove and a righthand 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 righthand 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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A box contains three pairs of blue gloves and two pairs of
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23 Jan 2019, 01:54
ashwini2k6jha wrote: VeritasKarishma wrote: arps wrote: A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a lefthand glove and a righthand 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 righthand 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 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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Re: A box contains three pairs of blue gloves and two pairs of
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26 Jan 2019, 13:05
‘Creating Analogy’ Denote Blueleft as ‘A’ , Blueright as ‘B’ , Greenleft as ‘C’ , and Greenright as ‘D’. Set formed [ AAA BBB CC DD ], with 10 data points. Q stem asks to extract 3 data points from the set such that we have one of the two pairs: AB or CD Possible combinations of AB: AB+ A or B or C or D Possible combinations of CD: CD+ A or B or C or D No. of ways of selecting each combo (divide by 2! to allow correction for repeats, wherever applicable): ABA= [3C1*3C1*2C1]/2! = 9 ABB= [3C1*3C1*2C1]/2! = 9 ABC= [3C1*3C1*2C1] = 18 ABD= [3C1*3C1*2C1]= 18 CDA= [2C1*2C1*3C1] = 12 CDB= [2C1*2C1*3C1] = 12 CDC= [2C1*2C1*1C1]/2! = 2 CDD= [2C1*2C1*1C1]/2! = 2 Favourable conditions: 9+9+18+18+12+12+2+2 = 82 Total conditions: 10C3 = 120 Probability= 82/120 = 41/60 Ans D



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Re: A box contains three pairs of blue gloves and two pairs of
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24 Jul 2019, 02:38
VeritasKarishma wrote: 12bhang wrote: 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 wrote: 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 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? 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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Re: A box contains three pairs of blue gloves and two pairs of
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