When four fair dice are rolled simultaneously, in how many possible outcomes does at least one of the dice show the number 3 on its upper face?

Introduction / Context:
This question tests the use of the complement principle in probability and counting. Counting the outcomes where at least one die shows a specific number is often easier by first counting all possible outcomes and then subtracting the outcomes where no die shows that number. Here, we focus on counting rather than probability, but the underlying logic is the same and very important for aptitude problems.

Given Data / Assumptions:
There are 4 fair dice rolled simultaneously. Each die has 6 faces numbered from 1 to 6. We are interested in outcomes where at least one die shows 3. Each die roll is independent of the others.

Concept / Approach:
The total number of outcomes when 4 dice are rolled is 6^4, since each die has 6 possible results. To find how many outcomes show at least one 3, it is easier to subtract from the total the number of outcomes where no die shows 3. If 3 is forbidden, each die has only 5 possible values (1, 2, 4, 5 or 6). Therefore, the number of outcomes with no 3 is 5^4. The required count is total outcomes minus outcomes with no 3.

Step-by-Step Solution:
Step 1: Compute the total number of outcomes when 4 dice are rolled. Step 2: Total outcomes = 6^4 = 6 * 6 * 6 * 6 = 1296. Step 3: Count outcomes where no die shows 3. Then each die has 5 options. Step 4: Outcomes with no 3 = 5^4 = 5 * 5 * 5 * 5 = 625. Step 5: Required outcomes with at least one 3 = total - no 3. Step 6: Compute 1296 - 625 = 671. Step 7: Therefore, there are 671 outcomes where at least one die shows the number 3.

Verification / Alternative check:
A direct method would try to count outcomes with exactly 1, 2, 3 or 4 dice showing 3 and sum those counts. Although possible, it is more tedious and prone to errors. For example, exactly one die showing 3 gives 4C1 * 5^3 outcomes, exactly two dice give 4C2 * 5^2, and so on. If you compute and sum these terms, you will again reach 671. This confirms that the complement method is both simpler and reliable.

Why Other Options Are Wrong:
The number 625 is the count of outcomes with no 3 at all, not the outcomes with at least one 3. The values 620 and 567 do not correspond to any standard combination or complement calculation in this context. The value 630 may arise from partial counting of only some cases. Only 671 agrees with the formula 6^4 - 5^4 and matches both the complement and expansion methods.

Common Pitfalls:
One common mistake is to misinterpret at least one as exactly one and count only one case. Another mistake is to forget to subtract from the total and instead add or multiply incorrectly. Some learners also miscalculate 6^4 or 5^4. Always break the problem into total outcomes and outcomes that do not satisfy the condition, then subtract. This complement idea appears frequently in exam questions involving dice, cards and other discrete random situations.

Final Answer:
The number of outcomes in which at least one die shows the number 3 is 671.