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

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Director
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If two different numbers are randomly selected from set { 1, [#permalink] New post 23 Jul 2010, 07:52
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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, 01:07, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Probability clarification [#permalink] New post 23 Jul 2010, 08:15
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] New post 23 Jul 2010, 08:46
you are right thanks!

the key is "different" numbers in the Q

i missed that
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Re: Probability clarification [#permalink] New post 23 Jul 2010, 12:06
Expert's post
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] New post 13 Dec 2012, 00:53
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] New post 10 Jan 2013, 20:54
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
Re: Probability clarification   [#permalink] 10 Jan 2013, 20:54
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