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Re: A box contains three pairs of blue gloves and two pairs of
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11 Jul 2017, 00:23
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, 00:43
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, 03:47
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, 18:49
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, 04:22
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, 00:54
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, 12:05
‘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




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