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Senior Manager  Status: Do and Die!!
Joined: 15 Sep 2010
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If three numbers are randomly selected from set A without  [#permalink]

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10 00:00

Difficulty:   55% (hard)

Question Stats: 68% (02:26) correct 32% (02:14) wrong based on 266 sessions

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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 10/7

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) => (1-1/7) * (1-1/6) => 5/7

where i'm going wrong.
Thanks

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Originally posted by shrive555 on 28 Dec 2010, 10:02.
Last edited by Bunuel on 07 Jul 2013, 05:52, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Math Expert V
Joined: 02 Sep 2009
Posts: 55609

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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) => (1-1/7) * (1-1/6) => 5/7

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 $$P=\frac{C^3_5}{C^3_7}=\frac{2}{7}$$.

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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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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.
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Re: If three numbers are randomly selected from set A without  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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) => (1-1/7) * (1-1/6) => 5/7

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 $$P=\frac{C^3_5}{C^3_7}=\frac{2}{7}$$.

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 = 1-1/7 = 6/7

What I am doing wrong?
Math Expert V
Joined: 02 Sep 2009
Posts: 55609
Re: If three numbers are randomly selected from set A without  [#permalink]

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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) => (1-1/7) * (1-1/6) => 5/7

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 $$P=\frac{C^3_5}{C^3_7}=\frac{2}{7}$$.

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 = 1-1/7 = 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}...
_________________ Re: If three numbers are randomly selected from set A without   [#permalink] 08 Jan 2019, 10:28
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