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# If two different numbers are randomly selected from a set

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Director
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If two different numbers are randomly selected from a set [#permalink]

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02 Jun 2005, 08:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If two different numbers are randomly selected from a set (1,2,3,4,5) what is the probability that their sum will be bigger than 4?
Senior Manager
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02 Jun 2005, 08:18
total possibilities: 5c2=10
you have 3 as a sum with:1+2
you have 4 as a sum with:1+3
hence
1-2/10=8/10=4/5
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02 Jun 2005, 10:04
I think we can have 2 possibilities for each -for eg for 2 we will have 1+2 and 2+1.Not sure though

total possibilities: 5c2=10
you have 3 as a sum with:1+2 and 2+1=2
you have 4 as a sum with:1+3 and 3+1=2
hence
1-4/10=6/10=3/5.

Thanks
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02 Jun 2005, 10:14
order doesn't matter!
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02 Jun 2005, 11:43
i also think it's 8/10=4/5

with 1 there are 2 sums more than 4
2- 3 sums
3- 2 sums
4- 1 sum

total 8 sums

8/c(5,2)=8/10=4/5
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02 Jun 2005, 21:53
Thanks for the replies, but i was thinking that the min sum is 3 and the max is 9 so we have 7 options in totatal. The prob that the sum is bigger than 4 is thus if we have sums 5,6,7,8,9 or 5/9
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03 Jun 2005, 09:23
BG wrote:
If two different numbers are randomly selected from a set (1,2,3,4,5) what is the probability that their sum will be bigger than 4?

I would go for 0.8

as there are 10 ways to pick two numbers from the list

two of those options (1,3 and 1,2) do not have a sum greater than four

thus probability of getting a sum greater than four is 8/10 or 0.8
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20 Dec 2007, 13:35
thearch wrote:
total possibilities: 5c2=10
you have 3 as a sum with:1+2
you have 4 as a sum with:1+3
hence
1-2/10=8/10=4/5

Alternatively,

you have 3 as a sum with: 1+2
1/5 * 2/4 = 1/10

you have 4 as a sum with:1+3
1/5 * 2/4 = 1/10

prob of either scenario = 2/10

1-2/10 = 4/5
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20 Dec 2007, 18:55
Guys, I think that the answer is 6/10 or 3/5:

What is the probability that their sum will be bigger than 4? = 1- probability that less than 4

We can have less that 4 with two ways:
1+3 = 3 + 1 = 4
1 + 2 = 2 + 1 = 3

i.e. probability = 4/(5C2) = 4/10

1 - 4/10 = 6/10 = 3/5
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20 Dec 2007, 21:29
BG wrote:
If two different numbers are randomly selected from a set (1,2,3,4,5) what is the probability that their sum will be bigger than 4?

what is the probability that the sum will be less or equal to four?

we have 1,2 and 2,1 1,3 and 3,1

1/5*1/4 *2 (b/c two ways to arrage 1 and 2)

same for 1 and 3

1/5*1/4*2

1/10+1/10=1/5 --> 4/5
Re: probability 2   [#permalink] 20 Dec 2007, 21:29
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