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81) From the even numbers between 1 and 9, two different

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81) From the even numbers between 1 and 9, two different [#permalink] New post 19 Jul 2008, 18:33
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81) From the even numbers between 1 and 9, two different even numbers are to be chosen at random. What is the probability that their sum will be 8?

is this ans not correct?
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v
total ways = 4c2.
total possible cases= 2(6,2 and 2,6).
so ans =2/6 = 1/3
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Re: prob gmat prep [#permalink] New post 19 Jul 2008, 18:41
First, I do not think the order here is important. You pick up two numbers at the same time and then add them. Therefore, count each combination as one .

Second, there are also other combination of numbers that can add upto 8
2,6
1,7
3,5

Thus Ans should be 3/6 =1/2..
Whats the OA.
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Re: prob gmat prep [#permalink] New post 19 Jul 2008, 18:44
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Total even numbers = 2,4,6,8

Total ways of selecting 2 out of 4 = 4C2 = 6 [this is only possible when order does not matter]

Total possibilities for 8 = (2,6) [order does not matter]

Probability = 1/6.

If you take order into account then total ways of ordering 2 out of 4 = 4P2 = 12
So as per your calculation answer should be 2/12 = 1/6
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Re: prob gmat prep [#permalink] New post 19 Jul 2008, 18:54
good explanation abhijit.
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Re: prob gmat prep [#permalink] New post 19 Jul 2008, 19:18
Like others clearly noted, only possiblity is 2 & 6
2/4 * 1/3 = 1/6
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Re: prob gmat prep [#permalink] New post 19 Jul 2008, 19:37
hmmm, the ans is 1/6. thanks guys
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Re: prob gmat prep [#permalink] New post 20 Jul 2008, 09:59
arjtryarjtry wrote:
hmmm, the ans is 1/6. thanks guys


if order matters, like you indicated with 2+6 or 6+2, then you can use permutations to get the same answer:

4!/(4-2)! = 12 possible permutations
2/12 = 1/6
Re: prob gmat prep   [#permalink] 20 Jul 2008, 09:59
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81) From the even numbers between 1 and 9, two different

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