Two fair six faced dice are rolled simultaneously. What is the probability that the sum of the numbers showing on the two upper faces is divisible by 3 or divisible by 4?

Introduction / Context:
This question combines divisibility with dice sums. When two dice are rolled, we can obtain sums from 2 to 12. We are asked for the probability that the sum is divisible by 3 or divisible by 4. This illustrates how to handle union probabilities in a finite sample space where events may overlap, because some sums are divisible by both 3 and 4.

Given Data / Assumptions:

• Two fair six sided dice are rolled simultaneously.
• Each die shows an integer from 1 to 6.
• There are 6 * 6 = 36 equally likely ordered outcomes.
• We are interested in sums that are multiples of 3 or multiples of 4.
• We must be careful not to double count sums that satisfy both conditions.

Concept / Approach:
We first list all possible sums from 2 to 12 and identify which sums are divisible by 3 or 4. We then count the number of ordered pairs of dice outcomes that yield these sums. Because each sum has a known number of generating pairs, this counting can be done by listing or by using known frequency patterns. The probability is the number of favorable outcomes divided by 36.

Step-by-Step Solution:
Step 1: The possible sums of two dice range from 2 to 12.Step 2: Identify sums divisible by 3: these sums are 3, 6, 9 and 12.Step 3: Identify sums divisible by 4: these sums are 4, 8 and 12.Step 4: The union of these sums is {3, 4, 6, 8, 9, 12}. Note that 12 appears in both lists.Step 5: Count the number of ordered outcomes giving each of these sums. The sum 3 has 2 outcomes, sum 4 has 3, sum 6 has 5, sum 8 has 5, sum 9 has 4 and sum 12 has 1, giving a total of 2 + 3 + 5 + 5 + 4 + 1 = 20 outcomes.Step 6: The probability that the sum is divisible by 3 or 4 is 20 / 36, which simplifies to 5 / 9 after dividing numerator and denominator by 4.
Verification / Alternative check:
We can verify our counts using the standard triangular distribution of two dice sums. The frequencies of sums from 2 to 12 are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2 and 1. Applying these counts to the sums 3, 4, 6, 8, 9 and 12 confirms the numbers used in the calculation. Another way is to write a small table of all 36 pairs and mark which meet the divisibility conditions, but that is more tedious. The consistency of the count with known sum patterns supports our result.

Why Other Options Are Wrong:
3/7 and 7/11 have denominators that do not equal 36 and do not naturally arise from any count of outcomes in this dice experiment. 6/13 is an odd fraction for this context and would correspond to 24 out of 52 if scaled, which is unrelated. 4/9 would be 16 favorable outcomes, but we have counted 20 outcomes that satisfy the given condition, making 5/9 the correct choice.

Common Pitfalls:
Students sometimes miscount the number of outcomes for each sum or forget to include sums that are divisible by both 3 and 4, such as 12. Another issue is accidentally counting each sum only once instead of counting all ordered pairs that yield that sum. It is important to distinguish between sums and the number of ways to achieve them when converting to probabilities.

Final Answer:
The probability that the sum is divisible by 3 or by 4 is 5/9.