A few ways of thinking about this:

1:: Since each ball is equally likely to be odd or even -- you have the same number of odds as you do of evens -- each sum is equally likely to be odd or even. So there's a 1/2 chance the sum is odd and a 1/2 chance it's even.

2:: Case by case, the three numbers we pick could be: Odd/Odd/Odd, Odd/Odd/Even, Odd/Even/Even, Even/Even/Even. The first and third of these (OOO and OEE) will have odd sums, the second and fourth will have even sums. Since the probability of three Odds = the probability of three Evens, and the probability of two Odds = the probability of two Evens, we again have equally likely outcomes.

That helps a lot, thank you!

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A box contains 100 balls, numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of three numbers on the balls selected from the box will be odd?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

==> Probability = specific cases of event/total number of cases
total number of cases:100*100*100
specific cases of event : e+e+o, e+o+e, o+e+e, or o+o+o are all the cases we have, and are 50.
Thus we have 4 cases of 50*50*50. The numerator is 50*50*50*4.
The probability that the addition of three numbers is odd is =50*50*50*4/100*100*100=1/2. Thus the answer is C.


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