varotkorn wrote:
MODIFIED QUESTION: A box contains 3,000 balls, numbered from 1 to 3,000. If NINETY NINE balls are selected at random and WITH replacement from the box, what is the probability that the sum of the NINETY NINE numbers on the balls selected from the box will be odd?
Q1. Will the answer to the above problem still be 1/2 because of symmetry in the number of odds and evens in the pool we pick?
Yes. If your sum is even after 98 selections, you need to pick an odd number, and if your sum is odd after 98 selections, you need to pick an even number. Either way, you always have a 1/2 probability of picking the "right" type of number with your 99th pick, so 1/2 is the answer.
varotkorn wrote:
Q2. If the question were WITHOUT replacement, is there a logical approach to solve to above modified question within time limit?
Yes, but I think that situation is probably too complicated for the GMAT. The answer is again 1/2;intuitively there should be no reason an odd or even sum should be more likely than the other if the set is half even, half odd. There are a few ways to prove that's the answer, though some rely on math beyond the scope of the test. Using only GMAT-level math, you could notice that these sequences have an equal probability of occurring:
EEE...EEE (all 99 selections are even) -- even sum
OOO...OOO (all 99 selections are odd) -- odd sum
and so do these sequences:
EEE...O... EEE (98 even selections, one odd) -- odd sum
OOO...E...OOO (98 odd selections, one even) -- even sum
and so on. So the sequences producing odd sums are balanced out by equally likely sequences producing even sums, and the answer is 1/2.
More abstractly, if you take any sequence of 99 Evens and Odds, and then just flip all the Even things to Odds, and all the Odds to Even, you'll get an equally likely sequence with the opposite type of sum (for example if you have a sequence with 46 evens and 53 odds, that produces an odd sum, but change all the even numbers to odd ones and vice versa, and now you get an even sum). So an even sum and odd sum must be equally likely.
Still, that question seems beyond GMAT scope, in part just because the numbers are so big that direct computational approaches become impossible. Usually in GMAT questions like this, there will be a 'fast' conceptual approach (like the ones I explained above) available, along with a slower-but-practical calculational approach (using ordinary probability rules). If you had a set of 10 numbers and 3 selections without replacement, that would be a more GMAT-like scenario.
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