2008 AMC 10B Problem 16

Split by the number of heads.

• Zero heads (probability $\tfrac14$): no die is rolled, the sum is $0$, which is even. Contributes $0$.
• One head (probability $\tfrac12$): one die, odd with probability $\tfrac12$. Contributes $\tfrac12 \cdot \tfrac12 = \tfrac14$.
• Two heads (probability $\tfrac14$): two dice. The sum is odd exactly when the dice have different parities, which happens with probability $\tfrac12$ (whatever the first die shows, the second matches its parity half the time). Contributes $\tfrac14 \cdot \tfrac12 = \tfrac18$.

Total:
$$0 + \frac14 + \frac18 = \frac38.$$

The answer is $\boxed{\textbf{(A)}\ \frac{3}{8}}$.

A two-stage experiment is handled by conditioning on the first stage and weighting each case by its probability. The useful fact is that a sum of any positive number of fair dice is odd exactly half the time (the last die decides the parity), so the only case that behaves differently is the degenerate one with no dice. The answer is simply $P(\text{at least one die}) \cdot \tfrac12 = \tfrac34 \cdot \tfrac12$.

• Treating the no-dice case as a coin flip too and answering 1/2, or forgetting that a sum of 0 counts as even.
• Weighting the three cases equally ($\tfrac13$ each) instead of $\tfrac14, \tfrac12, \tfrac14$.