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If three numbers are randomly selected from set A without
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Updated on: 07 Jul 2013, 04:52
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Set A: {1, 3, 4, 6, 9, 12, 15} If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3? A. 3/14 B. 2/7 C. 9/14 D. 5/7 E. 11/14 The sum of any three number from set B={3,6,12,15} would be divisible by 3. with combination method we can find how many ways we can select 3 numbers from set B. 5C3 = 10 so the probability would be 10/72nd method is of probability: The easy would be not selecting 1 and 4 from the set. so P(not 1 and not 4) = P (not 1) * P(not4) => (11/7) * (11/6) => 5/7Both Answers are wrong !! where i'm going wrong. Thanks
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Originally posted by shrive555 on 28 Dec 2010, 09:02.
Last edited by Bunuel on 07 Jul 2013, 04:52, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: probability/3
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28 Dec 2010, 09:17
shrive555 wrote: Set A: {1, 3, 4, 6, 9, 12, 15}
If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?
The sum of any three number from set B={3,6,12,15} would be divisible by 3. with combination method we can find how many ways we can select 3 numbers from set B. 5C3 = 10 so the probability would be 10/7
2nd method is of probability: The easy would be not selecting 1 and 4 from the set. so P(not 1 and not 4) = P (not 1) * P(not4) => (11/7) * (11/6) => 5/7
Both Answers are wrong !! where i'm going wrong. Thanks Set A: {1, 3, 4, 6, 9, 12, 15}
If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?A. 3/14 B. 2/7 C. 9/14 D. 5/7 E. 11/14 How can the probability be more than 1? Probability of an event = (# of favorable outcomes) / (total # of outcomes). There are 5 numbers you can choose from (3, 6, 9, 12, 15) so that the sum to be divisible by 3 and there are total of 7 numbers so \(P=\frac{C^3_5}{C^3_7}=\frac{2}{7}\). Answer: B.
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Re: probability/3
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28 Dec 2010, 09:31
ooo... i'm so dumb !! the denominator with 7C3. choosing 3 out of 7 but whats wrong with the 2nd method ?
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Re: probability/3
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Re: If three numbers are randomly selected from set A without
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20 May 2014, 08:11
sum of 3 multiples of 3 is divisible by 3. sum of 2 multiples of 3 with one non multiple of 3 will never be divisible by 3. sum of 1 multiple of 3 with 2 non multiples is only possible if sum of two non multiples is divisible by 3. 1,3,4,6,9,12,15 1 and 4 are non multiple. 5C3 ways to select multiples of 3. 4+1 = 5 thus third possibility one multiple of 3 with 2 non multiple is not possible. Therefor 5C3/7C3 = 2/7 ans.
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Re: If three numbers are randomly selected from set A without
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22 May 2014, 05:05
You can Get a multiple of 3 by adding multiples of 3 only...(Unless you have repeat numbers such as three 2's or 5 1's) 5 multiples of 3.. 5C3 Total possible combinations 7C3 = 2/7
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Re: If three numbers are randomly selected from set A without
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19 Sep 2017, 09:11
Bunuel wrote: shrive555 wrote: ooo... i'm so dumb !! the denominator with 7C3. choosing 3 out of 7 but whats wrong with the 2nd method ? I'm not sure what are doing in your second approach but using probability you can solve: 5/7*4/6*3/5=2/7. Bunuel how about reverse probability? this is what i tried and ended up with the wrong answer: \(1\frac{2}{7}*\frac{1}{6}*1+\frac{2}{7}*\frac{5}{6}*\frac{4}{5}= 1\frac{5}{21}\)



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Re: If three numbers are randomly selected from set A without
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20 Sep 2017, 22:54
No of ways of selecting 3 numbers out of the total of 7 numbers = 7C3 5 out of the total set of 7 numbers are multiples by 3 and hence, will account to the sum being divisible by 3 = No of ways of selecting favorable outcome = 5C3
P(E)  5C3/ 7C3 = 2/7 (B)



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Re: If three numbers are randomly selected from set A without
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23 Jun 2018, 21:54
The required probability is (5c3/7c3). The answer comes to 2/7. Option B it is !
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Re: If three numbers are randomly selected from set A without
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08 Jan 2019, 09:00
Bunuel wrote: shrive555 wrote: Set A: {1, 3, 4, 6, 9, 12, 15}
If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?
The sum of any three number from set B={3,6,12,15} would be divisible by 3. with combination method we can find how many ways we can select 3 numbers from set B. 5C3 = 10 so the probability would be 10/7
2nd method is of probability: The easy would be not selecting 1 and 4 from the set. so P(not 1 and not 4) = P (not 1) * P(not4) => (11/7) * (11/6) => 5/7
Both Answers are wrong !! where i'm going wrong. Thanks Set A: {1, 3, 4, 6, 9, 12, 15}
If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?A. 3/14 B. 2/7 C. 9/14 D. 5/7 E. 11/14 How can the probability be more than 1? Probability of an event = (# of favorable outcomes) / (total # of outcomes). There are 5 numbers you can choose from (3, 6, 9, 12, 15) so that the sum to be divisible by 3 and there are total of 7 numbers so \(P=\frac{C^3_5}{C^3_7}=\frac{2}{7}\). Answer: B. Hi Bunuel, If we want to do another way => 1  P(choosing 1 and 4) then as per the number of ways to chose 1 and 4 = (1,4,3), (1,4,6), (1,4,9), (1,4,12), and (1,4,15); in short 5C1 therefore P(choosing 1 and 4) should be = 5/(7c3) = 5/35 = 1/7 and P(NOT choosing 1 and 4) should be = 11/7 = 6/7 What I am doing wrong?



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Re: If three numbers are randomly selected from set A without
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08 Jan 2019, 09:28
harsh8686 wrote: Bunuel wrote: shrive555 wrote: Set A: {1, 3, 4, 6, 9, 12, 15}
If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?
The sum of any three number from set B={3,6,12,15} would be divisible by 3. with combination method we can find how many ways we can select 3 numbers from set B. 5C3 = 10 so the probability would be 10/7
2nd method is of probability: The easy would be not selecting 1 and 4 from the set. so P(not 1 and not 4) = P (not 1) * P(not4) => (11/7) * (11/6) => 5/7
Both Answers are wrong !! where i'm going wrong. Thanks Set A: {1, 3, 4, 6, 9, 12, 15}
If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?A. 3/14 B. 2/7 C. 9/14 D. 5/7 E. 11/14 How can the probability be more than 1? Probability of an event = (# of favorable outcomes) / (total # of outcomes). There are 5 numbers you can choose from (3, 6, 9, 12, 15) so that the sum to be divisible by 3 and there are total of 7 numbers so \(P=\frac{C^3_5}{C^3_7}=\frac{2}{7}\). Answer: B. Hi Bunuel, If we want to do another way => 1  P(choosing 1 and 4) then as per the number of ways to chose 1 and 4 = (1,4,3), (1,4,6), (1,4,9), (1,4,12), and (1,4,15); in short 5C1 therefore P(choosing 1 and 4) should be = 5/(7c3) = 5/35 = 1/7 and P(NOT choosing 1 and 4) should be = 11/7 = 6/7 What I am doing wrong? You are missing cases there. What about: {1, 3, 6}; {1, 3, 9}; {1, 3, 12}; {1, 3, 15}; {1, 6, 9}; {1, 6, 12}; {1, 6, 15}; {1, 9, 12}; {1, 9, 15}...
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