From the even numbers between 1 and 9, two different even : Quant Question Archive [LOCKED]
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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]

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12 Dec 2003, 06:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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Regarding even number between 1 and 9 [#permalink]

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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
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Posts: 424
Location: Aus
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12 Dec 2003, 07:37
I am getting 1/6.

Do we have to count (2,6) twice ?
Manager
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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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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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12 Dec 2003, 18:42
Can this be solved using combinations?

2C2 / 4C2 = 1/6 ...
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12 Dec 2003, 23:13

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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