probability

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ksb214: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Thu Jun 18, 2009 12:00 pm

by ksb214 » Sun Jun 28, 2009 3:38 pm

Can you help me in solving this problem from Gmac ?

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Source: Beat The GMAT — Problem Solving | Sun Jun 28, 2009 3:38 pm

GMATQuantCoach: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Jun 28, 2009 8:58 am; Location: Online; Thanked: 13 times; GMAT Score:51

by GMATQuantCoach » Sun Jun 28, 2009 10:47 pm

For the sum to be odd, the three balls must be odd,odd,odd, or even,even,odd, or even, odd, even, or odd, even, even

The probability of each case is 1/8 ( 1/2 * 1/2 * 1/2).

Then total probability is 4 * 1/8 = 1/2

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Tue Jun 30, 2009 9:54 am

Alternatively: it doesn't matter what your first two selections are. They either add to an even number, in which case you need the third ball to be odd (probability 1/2) or they add to an odd number, in which case you need the third ball to be even (probability 1/2). No matter what the situation after your first two selections, there's a 1/2 probability you'll arrive at an odd sum after your third selection.

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gmat_2010: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Thu Apr 09, 2009 10:33 am

by gmat_2010 » Thu Jul 30, 2009 7:16 am

Yup, I arrived at the solution the same way Ian did. Probability of any ball being odd is same of that being even. And the total sum can be only either even or odd. Hence, there is no preference for even or odd. Hence both are equally likely. Exhaustive set. Hence, 1/2 for each.

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