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Two distinct numbers a and b are chosen randomly from the set of first
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03 Feb 2015, 09:45
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Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3? A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190 Kudos for a correct solution.
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Re: Two distinct numbers a and b are chosen randomly from the set of first
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09 Feb 2015, 04:50
Bunuel wrote: Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3?
A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Any number could be written in the form of one among the following: (i) 3k (ii) 3k + 1 (iii) 3k + 2 Only if the 2 numbers are of different types would a – b not be divisible by 3: Numbers of type 3k > 3, 6, 9, ......... 18 > Total 6 numbers Numbers of type 3k + 1 > 1, 4, 7, .... 19 > Total 7 numbers Numbers of type 3k + 2 > 2, 5, 8, 11 ...20 >Total 7 numbers Total possible outcomes = 20 * 19 = 380 Favorable Cases: 1) Picking from Type 1 and Type 2: 2 * (6 * 7) = 84 2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84 3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98 Hence, the required probability = (84 + 84 + 98) / 380, and E is the correct answer.
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Re: Two distinct numbers a and b are chosen randomly from the set of first
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03 Feb 2015, 13:20
Here is what i thought:
From 20 first positive integers, we get that 6 of them are divisible by 3. For the equation not to be divisible by 3, we have the following possible combinations:
both of the numbers not being divisible by 3:
14/20 * 13/19 = 91/190
one of them is divisible by 3 and the other isn't:
14/20 * 6/19 = 42/190
because we need combination 1 OR combination 2 to achieve the requested, we need to sum both fractions:
91/190 + 42/190 = 133/190
For me, answer E. i guess i forgot something and made it completly wrong.




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Re: Two distinct numbers a and b are chosen randomly from the set of first
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03 Feb 2015, 13:31
Firstly, The pattern is 1 4 7 10 13 16 19 = Pick two numbers from 7 > 7p2= 42 2 5 8 11 14 17 20 = pick two number from 7 > 7p2= 42 3 6 9 12 15 18= Pick two numbers from 6> 6P2= 30. Total permuations that are divisible by 3 are 114. Total permutions are 20p2= 190. 1 114/190. This is exactly where I am struck. I am getting a strange answer. I guess, I have to wait for someone to crack this one. \\ Bunuel please guide me what I am doing wrong. Bunuel wrote: Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3?
A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190
Kudos for a correct solution.
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Kudos to you, for helping me with some KUDOS.



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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03 Feb 2015, 17:50
I think you just made the mistake that 20P2 will be 20 * 19 = 380 and NOT 190.. If you consider this, that would make 114/380 (probability that the no is divisible by 3) Subtract this probability from 1, 1 114/380 = 266/380 = 133/190 Ans (E) shriramvelamuri wrote: Firstly,
The pattern is
1 4 7 10 13 16 19 = Pick two numbers from 7 > 7p2= 42
2 5 8 11 14 17 20 = pick two number from 7 > 7p2= 42
3 6 9 12 15 18= Pick two numbers from 6> 6P2= 30.
Total permuations that are divisible by 3 are 114.
Total permutions are 20p2= 190.
1 114/190.
This is exactly where I am struck. I am getting a strange answer. I guess, I have to wait for someone to crack this one. \\



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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03 Feb 2015, 18:03
I started to write the numbers out. Eventually, I thought that since a little less than a third of the numbers are divisible by three, the probability of the absolute value of any number minus another number not being divisible by three would be a little more than 2/3, so I went with answer E.



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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22 Feb 2015, 12:20
Bunuel wrote: Bunuel wrote: Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3?
A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Any number could be written in the form of one among the following: (i) 3k (ii) 3k + 1 (iii) 3k + 2 Only if the 2 numbers are of different types would a – b not be divisible by 3: Numbers of type 3k > 3, 6, 9, ......... 18 > Total 6 numbers Numbers of type 3k + 1 > 1, 4, 7, .... 19 > Total 7 numbers Numbers of type 3k + 2 > 2, 5, 8, 11 ...20 >Total 7 numbers Total possible outcomes = 20 * 19 = 380 Favorable Cases: 1) Picking from Type 1 and Type 2: 2 * (6 * 7) = 84 2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84 3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98 Hence, the required probability = (84 + 84 + 98) / 380, and E is the correct answer. Hi Bunuel, Why exactly are you multiplying by 2 when you are calculating the favourable cases. Many thanks



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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22 Feb 2015, 13:20
bhatiavai wrote: Bunuel wrote: Bunuel wrote: Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3?
A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Any number could be written in the form of one among the following: (i) 3k (ii) 3k + 1 (iii) 3k + 2 Only if the 2 numbers are of different types would a – b not be divisible by 3: Numbers of type 3k > 3, 6, 9, ......... 18 > Total 6 numbers Numbers of type 3k + 1 > 1, 4, 7, .... 19 > Total 7 numbers Numbers of type 3k + 2 > 2, 5, 8, 11 ...20 >Total 7 numbers Total possible outcomes = 20 * 19 = 380 Favorable Cases: 1) Picking from Type 1 and Type 2: 2 * (6 * 7) = 84 2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84 3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98 Hence, the required probability = (84 + 84 + 98) / 380, and E is the correct answer. Hi Bunuel, Why exactly are you multiplying by 2 when you are calculating the favourable cases. Many thanks 1) a is from Type 1 and b is from Type 2 + b is from Type 1 and a is from Type 2... The same for (2) and (3).
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Re: Two distinct numbers a and b are chosen randomly from the set of first
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27 Feb 2015, 11:50
1. Pick 2 from {1,2, ..... 20} 20*19 = 380 2. figure out the probability that a–b would be divisible by 3 1) {3, 6, 9, 12, 15, 18 } 6*5 = 30 2) {1,4,7,10,13, 16, 19} 7*6 = 42 3) {2, 5, 8, 11, 14, 17, 20} 7*6 = 42 3. the probability that a–b would not be divisible by 3 = (380  304242) / 380 = 133/190
So the correct answer is E



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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09 Aug 2015, 08:13
shriramvelamuri wrote: Firstly, The pattern is 1 4 7 10 13 16 19 = Pick two numbers from 7 > 7p2= 42 2 5 8 11 14 17 20 = pick two number from 7 > 7p2= 42 3 6 9 12 15 18= Pick two numbers from 6> 6P2= 30. Total permuations that are divisible by 3 are 114. Total permutions are 20p2= 190. 1 114/190. This is exactly where I am struck. I am getting a strange answer. I guess, I have to wait for someone to crack this one. \\ Bunuel please guide me what I am doing wrong. Bunuel wrote: Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3?
A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190
Kudos for a correct solution. when you select 2 numbers from a set of numbers, you use combination and not permeutation (one of the very common mistakes), so you should do = \(1  (7C2 + 7C2 + 6C2)/ 20C2\)



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Two distinct numbers a and b are chosen randomly from the set of first
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10 Aug 2015, 06:05
I have handled this one with no real mathematics involved.
For each number A selected you have 19 possible results for ab, because when you select A you have 19 numbers remainding.
You know that only 6 results of ab will give you a number divisible by 6 ( 3, 6, 9, ...18) . Then, 14 results out of 19 won't be divisible by 3.
Look, we've just said that 14 won't be divisible by 3 out of 19 possibles outcomes... 133/190 is approximatively this number.



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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10 Aug 2015, 06:29
Bunuel wrote: Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3?
A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190 t
Kudos for a correct solution. Let find the probability that ab is divisible by 3. Then ab has to be 3, 6, 9, 12, 15, 18 (cannot be 0 since a and b are distinct) To get ab as 3, ab could be 41, 52.....2017 => 17 Times. But since this is modulus, 41 = 14. Hence to get 3, there are 34 ways. Similarly to get 6 => 14*2 = 28 9 => 11*2 = 22 12 => 8*2 = 16 15 => 5*2 = 10 18 => 2*2 = 4 Hence total ways to get ab divisible by 3 = 114 Total Outcomes = 20 x 19 = 380 Probability to get ab divisible by 3 is 114 / 380 => 57/190 Probability that ab is not divisible by 3 is = 1  (57/190) = 133/190 Option E  Please give kudos if you like this post



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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16 Nov 2015, 10:08
Bunuel wrote: Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that a–b would not be divisible by 3?
A. 49/380 B. 21/95 C. 49/190 D. 42/95 E. 133/190
Kudos for a correct solution. I have a better way to do this problem: number 120 we have 19*20= 380 ways to choose a pair. from 120 each number have 6 ways to choose a number that ab divisible by 3: EX 20 must subtract by 17,14,11,8,5,3 to divisible by 3: so from 120 we have 20*6= 120 ways to choose, but we didnt want a same number, hence 1206= 114 so the percentage is 114/380 to choose a pair that divisible by 3, ===> hence the chance to pick a pair that not divisible by 3 is 1 (114/380)= 266/380 =133/190 and this is final answer.



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Re: Two distinct numbers a and b are chosen randomly from the set of first
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12 Feb 2018, 13:59
Hi All, The answer choices to this question provide a great 'logic shortcut' that you can take advantage of. The calculation that we're asked to consider (AB) and the limited possible values for the two distinct variables (120, inclusive) means that we're dealing with possible answers that are consecutive integers: 1 through 19, inclusive. When dealing with consecutive integers, only 1 out of every "set" of 3 will be divisible by 3. This question asks for the probability that the calculation will NOT be divisible by 3. Thus, the answer MUST be greater than 1/2 (closer to about 2/3). There's ONLY one answer that's greater than one half.... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Two distinct numbers a and b are chosen randomly from the set of first
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25 Feb 2019, 11:05
ab can have 19 values from 1 to 19 and out of these 19 values only 6 i.e. 3,6,9,12,15 and 18 are divisible by 3. Thus, 13 values are "not" divisible by 3. The probability can is thus 13/19 or 130/190. A close approximate to 133/190. In any case, the probability should be more than 1/2. Only option E is >1/2




Re: Two distinct numbers a and b are chosen randomly from the set of first
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