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

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Manager
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From the even numbers between 1 and 9, two different even [#permalink] New post 12 Dec 2003, 06:20
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?


My answer is

Probability = No. of favourable outcomes/ no. of possible outcomes
= 4/8
=1/2

But there is a better answer. Can anyone point out the proper workings?
Intern
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Regarding even number between 1 and 9 [#permalink] New post 12 Dec 2003, 07:31
For even numbers between 1 and 9 they are 2, 4, and 6. So the number of favourable outcomes would be {(2,6) (6,2)} and the total number of possible outcomes {(2,4) (4,2) (2,6) (6,2) (4,6) (6,4)}

2/6=1/3.

Please correct me if I am wrong.

How did you get 1/2 ? Can you esplain please.
-------------------------------------------------------------------------
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?


My answer is

Probability = No. of favourable outcomes/ no. of possible outcomes
= 4/8
=1/2
Senior Manager
Senior Manager
User avatar
Joined: 05 May 2003
Posts: 427
Location: Aus
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GMAT Tests User
 [#permalink] New post 12 Dec 2003, 07:37
I am getting 1/6.

Do we have to count (2,6) twice ? :?
Senior Manager
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 [#permalink] New post 12 Dec 2003, 14:51
I get 1/6

even numbers between 1 and 9 = 2,4,6,8
total outcomes = 6
how you can get 8 = 2 + 6 .. this is one .. isn't it the same as 6 and 2??

if 2,6 is different from 6,2 then the prob = 2/6 = 1/3
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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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GMAT Tests User
 [#permalink] New post 12 Dec 2003, 17:14
pitts20042006 wrote:
I get 1/6

even numbers between 1 and 9 = 2,4,6,8
total outcomes = 6
how you can get 8 = 2 + 6 .. this is one .. isn't it the same as 6 and 2??

if 2,6 is different from 6,2 then the prob = 2/6 = 1/3


If you are going to make the order count in the "successes" you have to also make it count in the "possibilities".

Either way, the number should come out to 1/6.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
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 [#permalink] New post 12 Dec 2003, 18:42
Can this be solved using combinations?

2C2 / 4C2 = 1/6 ...
Manager
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 [#permalink] New post 12 Dec 2003, 23:13
The correct answer is 1/6.

Am I missing somewhere?

No. of favourable outcomes
= {(6,2), (4,4)}
= 2

No. of possible outcomes
= {(2,2),(2,4),(2,6),(2,8),(4,4),(4,6),(4,8),(6,6)(6,8),(8,8)}
= 10
  [#permalink] 12 Dec 2003, 23:13
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From the even numbers between 1 and 9, two different even

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