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Director
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If two different numbers are randomly selected from a set [#permalink]
 02 Jun 2005, 09:08
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?
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Director
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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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Intern
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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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Director
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order doesn't matter!
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Manager
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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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Director
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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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Senior Manager
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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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CEO
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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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Manager
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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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CEO
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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
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