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

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Manager
Joined: 28 Apr 2003
Posts: 95
Location: Singapore
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From the even numbers between 1 and 9, two different even [#permalink]  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?

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
Joined: 11 Dec 2003
Posts: 49
Location: IN
Followers: 1

Kudos [?]: 9 [0], given: 0

Regarding even number between 1 and 9 [#permalink]  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?

Probability = No. of favourable outcomes/ no. of possible outcomes
= 4/8
=1/2
Senior Manager
Joined: 05 May 2003
Posts: 426
Location: Aus
Followers: 2

Kudos [?]: 5 [0], given: 0

I am getting 1/6.

Do we have to count (2,6) twice ?
Senior Manager
Joined: 12 Oct 2003
Posts: 251
Location: USA
Followers: 1

Kudos [?]: 5 [0], given: 0

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
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 71 [0], given: 0

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
Joined: 06 Jun 2003
Posts: 59
Followers: 1

Kudos [?]: 1 [0], given: 0

Can this be solved using combinations?

2C2 / 4C2 = 1/6 ...
Manager
Joined: 28 Apr 2003
Posts: 95
Location: Singapore
Followers: 1

Kudos [?]: 0 [0], given: 0

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
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