In a single throw of two fair six faced dice, what is the probability that neither a doublet (both dice showing the same number) nor a total of 8 appears on the upper faces?

Introduction / Context:
This problem combines two conditions in a dice experiment. We must avoid both a doublet and a sum of 8 when two dice are thrown. Such questions reinforce understanding of how to handle combined events and how to apply the idea of excluding unwanted outcomes from the full sample space of 36 possibilities.

Given Data / Assumptions:

• Two fair six sided dice are thrown together.
• Each die can show a number from 1 to 6.
• All 36 ordered pairs of outcomes are equally likely.
• A doublet is any outcome where both dice show the same number, such as (2, 2).
• The sum 8 consists of all ordered pairs whose total equals 8.

Concept / Approach:
We find the probability of the complement event by subtracting the probability of undesirable outcomes from 1. The undesirable outcomes are outcomes that are doublets, or that give a total of 8, or both. We use the union of events formula, which subtracts the overlap so that shared outcomes are not counted twice. Then we subtract this union probability from 1 to get the desired probability.

Step-by-Step Solution:
Step 1: Total possible outcomes from rolling two dice are 6 * 6 = 36.Step 2: Count the number of doublets. These are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6), giving 6 outcomes.Step 3: Count the outcomes that give a sum of 8. The valid ordered pairs are (2,6), (3,5), (4,4), (5,3) and (6,2), giving 5 outcomes.Step 4: Identify the overlap, that is outcomes that are both doublets and sum to 8. The only such outcome is (4,4).Step 5: Use the union count formula: total undesirable outcomes = 6 + 5 minus 1 = 10.Step 6: Probability of an undesirable outcome is 10 / 36, so the desired probability is 1 minus 10 / 36 = 26 / 36, which simplifies to 13 / 18.
Verification / Alternative check:
We can double check by listing the 10 undesirable outcomes explicitly and confirming that they are distinct and correctly counted. The doublets are 6 in number, and among the five sum 8 outcomes, only one overlaps. Removing that single common case from the total of 11 distinct descriptions gives 10 unique undesirable outcomes. Dividing 10 by 36 gives a probability of 5/18 for the union, so its complement is indeed 13/18. Both ways lead to the same result, confirming correctness.

Why Other Options Are Wrong:
7/15 and 5/18 come from incorrect counting or misinterpretation of the union and intersection. 3/16 is far from any fraction with denominator 36 and does not correspond to a natural count. 11/18 would result from subtracting 7/36 instead of 10/36, which would clearly overlook some outcomes.

Common Pitfalls:
A frequent mistake is to add the counts of doublets and sum 8 outcomes without subtracting the overlap, which double counts the outcome (4,4). Another error is to misidentify the pairs that sum to 8, perhaps including impossible pairs such as (1,7). Remember that each die only has faces from 1 to 6, and that union counts require careful attention to shared outcomes.

Final Answer:
The probability that neither a doublet nor a total of 8 appears is 13/18.