Two fair dice are thrown simultaneously. What is the probability that one die shows an even number and the other die shows an odd number?

Introduction / Context:
This question tests basic probability with two dice and focuses on parity, that is, whether numbers are even or odd. When two fair dice are rolled, each die independently shows one of six outcomes. You must find the probability that exactly one die is even and the other is odd, a very standard pattern in aptitude problems.

Given Data / Assumptions:

• Two fair six sided dice are thrown at the same time.
• Each die has faces numbered 1, 2, 3, 4, 5, 6.
• Even numbers on a die are 2, 4, 6 (three outcomes).
• Odd numbers on a die are 1, 3, 5 (three outcomes).
• We want one die to be even and the other to be odd.

Concept / Approach:
Since the dice are independent, we can compute the probability for each die and then apply the multiplication rule. There are two favorable patterns:

• First die even and second die odd.
• First die odd and second die even.

The total probability of exactly one even and one odd is the sum of the probabilities of these two disjoint events.

Step-by-Step Solution:
Step 1: Probability that a single die shows an even number = 3/6 = 1/2. Step 2: Probability that a single die shows an odd number = 3/6 = 1/2. Step 3: Probability that the first die is even and the second die is odd = (1/2) * (1/2) = 1/4. Step 4: Probability that the first die is odd and the second die is even = (1/2) * (1/2) = 1/4. Step 5: These two cases are mutually exclusive, so add them: 1/4 + 1/4 = 1/2. Step 6: Therefore, the probability of getting one even number and one odd number is 1/2.

Verification / Alternative check:
You can also solve this by counting outcomes in the full sample space. There are 6 * 6 = 36 possible ordered pairs for two dice. For each even value on the first die (2, 4, 6), the second die can be any of the 3 odd values, giving 3 * 3 = 9 outcomes of the type (even, odd). Similarly, for each odd value on the first die (1, 3, 5), the second die can be any of the 3 even values, again giving 3 * 3 = 9 outcomes of the type (odd, even). So favorable outcomes = 9 + 9 = 18. Probability = 18 / 36 = 1/2, which agrees with the earlier method.

Why Other Options Are Wrong:
Option 1/3 and option 1/6 both underestimate the true probability and would correspond to incorrect counting of favorable cases. Option 2/3 and option 5/6 are too large and might result from including outcomes where both numbers are even or both are odd. None of these incorrect values match either the stepwise calculation or the exhaustive counting method, both of which clearly give 1/2.

Common Pitfalls:
One common mistake is to confuse exactly one even and one odd with at least one even or at least one odd. Another mistake is to forget that the two patterns (even, odd) and (odd, even) are distinct events that should both be counted. Some students also attempt to count outcomes by inspection but miss several pairs, leading to an incorrect numerator. Writing out the sample space pattern or using symmetry and independence properly can prevent these errors.

Final Answer:
The probability of getting one even number and one odd number when two dice are thrown is 1/2.