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by ksb214 » Sun Jun 28, 2009 3:38 pm
Can you help me in solving this problem from Gmac ?
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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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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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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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