A box holds 12 tickets: 3 of them are marked 1, five are marked 2 and four are marked 3. A ticket is drawn at random, its number is noted and it is PUT BACK; a second ticket is then drawn. What is the probability that the two numbers add up to an ODD total?

A total is odd exactly when one ticket is odd and the other is even, so only the parity of each draw matters. The odd tickets are the 3 ones and the 4 threes, so $P$(odd) = 7/12, and $P$(even) = 5/12 from the five twos. The ticket is replaced, so the draws are independent, and an odd total arises in two orders: $P=2\times (7/12)(5/12)=70/144=35/72$. The value 35/144 keeps only one of those two orders, odd then even, and forgets even then odd. The value 37/72 is the complement, an even total, which comes from $(7/12{)}^2+(5/12{)}^2=74/144$; as a check, 35/72 + 37/72 = 1. The value 1/2 assumes odd and even totals must balance, which would need the odd and even tickets to be equally many — here they are 7 against 5.