Hi,
Each of the 3 balls picked will be either even or odd. So, total cases will be 2^3. In each take picking an even is as likely as picking an odd.
For sum to be odd,
1)all 3 odd -> 1
2)2 even, 1odd -> 3
So, probability is (1+3)/8 = 1/2.

My solution (I was solving this on a timed drill so, let me take you through what I did):
Step 1: I started listing cases as all possible cases (I thought at the time) will be equal to 6 (3!=3*2*1)
So the cases:
OOO
EEE
OOE
EOO
EEO
OEE

After listing these I realized there were two more cases:
EOE
OEO

Step 2: Start figuring out which case will lead to a favorable scenario i.e. an odd number.
OOO = O
EEE = E
OOE = E
EOO = E
EEO = O
OEE = O

EOE = O
OEO = E

Final Step: So we have total 8 cases and 4 favorable cases.
Ans: 4/8 = 1/2

After, solving I also realized that the rule which says, "the probability of a ball picked at random and with replacement from a pool of balls is equal to the probability of the first ball to be picked as a favorable scenario" can also be applied, if I am not mistaken. Can any of the experts, please, clarify?