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# If two different numbers are randomly selected from set { 1,

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If two different numbers are randomly selected from set { 1, [#permalink]

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23 Jul 2010, 08:52
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If two different numbers are randomly selected from set { 1, 2, 3, 4, 5} what is the probability that the sum of the two numbers is greater than 4?

OA ia
[Reveal] Spoiler:
4/5

[Reveal] Spoiler:
My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

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Last edited by Bunuel on 13 Dec 2012, 02:07, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Probability clarification [#permalink]

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23 Jul 2010, 09:15
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rxs005,
First problem is 5C2 = 10 .. Not 20..

Point 2, you sum should be greater than 4, so these are the following possibilities you have
({1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}) which leads to 8 pairs

So the probability is 8 on 10 or 4/5..
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Re: Probability clarification [#permalink]

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23 Jul 2010, 09:46
you are right thanks!

the key is "different" numbers in the Q

i missed that
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Re: Probability clarification [#permalink]

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23 Jul 2010, 13:06
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rxs0005 wrote:
If two different numbers are randomly selected from set { 1, 2, 3, 4, 5} what is the probability that the sum of the two numbers is greater than 4?

OA ia 4/5

My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

Another approach: $$P(of \ event \ X)=1-P(of \ opposite \ event \ X)$$. Opposite event would be if we choose 2 different number so that their sum will be less or equal to 4.

# of total outcomes is $$C^2_5=10$$ (total # of choosing 2 different numbers from the set of 5 different numbers);

# of outcomes when $$sum\leq{4}$$ is (1,2) and (1,3), so 2.

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

Answer: $$\frac{4}{5}$$.

P. S. In the future pleas post the answer choices too.
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Re: Probability clarification [#permalink]

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13 Dec 2012, 01:53
sridhar wrote:
rxs005,
First problem is 5C2 = 10 .. Not 20..

Point 2, you sum should be greater than 4, so these are the following possibilities you have
({1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}) which leads to 8 pairs

So the probability is 8 on 10 or 4/5..

How is the pare 1,4 GREATER than 4, I would say it is equal to four and therefore the answer should be 7/10.
Any argument ?
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Re: Probability clarification [#permalink]

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10 Jan 2013, 21:54
arnijon90 wrote:
sridhar wrote:
rxs005,
First problem is 5C2 = 10 .. Not 20..

Point 2, you sum should be greater than 4, so these are the following possibilities you have
({1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}) which leads to 8 pairs

So the probability is 8 on 10 or 4/5..

How is the pare 1,4 GREATER than 4, I would say it is equal to four and therefore the answer should be 7/10.
Any argument ?

Because: 1+4 = 5 and 5>4
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Re: If two different numbers are randomly selected from set { 1, [#permalink]

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04 Jun 2016, 21:43
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Re: If two different numbers are randomly selected from set { 1, [#permalink]

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10 Jun 2016, 22:24
Bunuel wrote:
rxs0005 wrote:
If two different numbers are randomly selected from set { 1, 2, 3, 4, 5} what is the probability that the sum of the two numbers is greater than 4?

OA ia 4/5

My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

Another approach: $$P(of \ event \ X)=1-P(of \ opposite \ event \ X)$$. Opposite event would be if we choose 2 different number so that their sum will be less or equal to 4.

# of total outcomes is $$C^2_5=10$$ (total # of choosing 2 different numbers from the set of 5 different numbers);

# of outcomes when $$sum\leq{4}$$ is (1,2) and (1,3), so 2.

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

Answer: $$\frac{4}{5}$$.

P. S. In the future pleas post the answer choices too.

Hi Bunuel,

Why not choosing (1,2) (2,1) (3,1) and (1,3)?

Ravi
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Re: If two different numbers are randomly selected from set { 1, [#permalink]

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18 Jun 2016, 11:24
gmatravi wrote:
Bunuel wrote:
rxs0005 wrote:
If two different numbers are randomly selected from set { 1, 2, 3, 4, 5} what is the probability that the sum of the two numbers is greater than 4?

OA ia 4/5

My approach was

total outcomes 5 C 2 = 20

outcomes where sum > 4 are

1,4 1,5 = 2
2,3 2,4 2,5 = 3
3,2, 3,3 3,4 3,5 = 4

4 any of the 5 = 5
5 any of the 5 = 5

total 19

so P(>4) = 19/20

why is this wrong

Another approach: $$P(of \ event \ X)=1-P(of \ opposite \ event \ X)$$. Opposite event would be if we choose 2 different number so that their sum will be less or equal to 4.

# of total outcomes is $$C^2_5=10$$ (total # of choosing 2 different numbers from the set of 5 different numbers);

# of outcomes when $$sum\leq{4}$$ is (1,2) and (1,3), so 2.

$$P=1-\frac{2}{10}=\frac{4}{5}$$.

Answer: $$\frac{4}{5}$$.

P. S. In the future pleas post the answer choices too.

Hi Bunuel,

Why not choosing (1,2) (2,1) (3,1) and (1,3)?

Ravi

I am having the same doubt as you, can someone enlighten our path?
Re: If two different numbers are randomly selected from set { 1,   [#permalink] 18 Jun 2016, 11:24
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# If two different numbers are randomly selected from set { 1,

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