A six-faced die is biased so that an even number is twice as likely as an odd number. It is thrown twice. What is the probability that the sum of the two outcomes is even?

Introduction / Context:
The die favors even results. The sum of two throws is even if both outcomes are even or both are odd. We must first compute the marginal probabilities of even and odd under the bias, then combine them for two throws (independent trials).

Given Data / Assumptions:

• Even outcome is twice as likely as an odd outcome.
• There are 3 even faces (2,4,6) and 3 odd faces (1,3,5).
• All even faces share equal probability; all odd faces share equal probability.
• Throws are independent.

Concept / Approach:
Assign weight 2 to each even face and weight 1 to each odd face. Total weight = 3×2 + 3×1 = 9. Thus P(single even) = 6/9 = 2/3 and P(single odd) = 3/9 = 1/3. For two throws, P(sum even) = P(both even) + P(both odd) = (2/3)^2 + (1/3)^2.

Step-by-Step Solution:

Verification / Alternative check:
Because the die is symmetric within parity classes, only parity probabilities matter, not specific face values; independence across throws validates the product rule.

Why Other Options Are Wrong:
1/12, 1/6, and 1/3 ignore the elevated chance of evens; the correct value exceeds 1/2 because evens are frequent.

Common Pitfalls:
Assigning double probability to the set of evens but forgetting there are three even faces (leading to mis-normalization).

Final Answer:
5/9