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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?

My answer is

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

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

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