In a simultaneous throw of a pair of fair six faced dice, what is the probability that the total (sum) of the numbers showing on the two upper faces is strictly greater than 7?

Introduction / Context:
This problem asks for the probability that the sum of two dice exceeds 7. Sums obtained when rolling two fair dice range from 2 to 12. The sums greater than 7 are 8, 9, 10, 11 and 12. This classic probability question helps students become comfortable with counting outcomes from the two dice sum distribution.

Given Data / Assumptions:

• Two fair six sided dice are rolled at the same time.
• Each die can show an integer from 1 to 6.
• There are 6 * 6 = 36 equally likely ordered outcomes.
• We need the probability that the sum of the outcomes is strictly greater than 7.

Concept / Approach:
We count how many ordered pairs (first die, second die) produce sums of 8, 9, 10, 11 or 12. Using the known distribution of sums, we can quickly determine these counts. The probability is then calculated as the number of favorable outcomes divided by 36. This uses basic counting and the idea of equally likely outcomes in a finite sample space.

Step-by-Step Solution:
Step 1: The total number of ordered outcomes when rolling two dice is 36.Step 2: Identify sums greater than 7, which are 8, 9, 10, 11 and 12.Step 3: Use the standard sum distribution: sum 8 has 5 outcomes, sum 9 has 4 outcomes, sum 10 has 3 outcomes, sum 11 has 2 outcomes and sum 12 has 1 outcome.Step 4: Add these counts to find the total number of favorable outcomes: 5 + 4 + 3 + 2 + 1 = 15.Step 5: The required probability is favorable outcomes divided by total outcomes, so probability = 15 / 36.Step 6: Simplify 15 / 36 by dividing numerator and denominator by 3 to get 5 / 12.
Verification / Alternative check:
As an alternative, we can use the complement event. The complement of the sum being greater than 7 is the sum being less than or equal to 7. The sums 2, 3, 4, 5, 6 and 7 have 1, 2, 3, 4, 5 and 6 outcomes respectively, totaling 21 outcomes. Thus, the probability of sum less than or equal to 7 is 21 / 36, so the probability of sum greater than 7 is 1 minus 21 / 36 = 15 / 36, which again simplifies to 5 / 12. This matches our earlier result.

Why Other Options Are Wrong:
1/2 would correspond to 18 favorable outcomes, which is more than the actual count of 15. 7/15 has denominator 15 and does not fit the 36 outcome space. 3/12 simplifies to 1/4, implying only 9 favorable outcomes, which is too few. 11/36 would correspond to 11 favorable outcomes, again not matching the counted 15 outcomes for sums greater than 7.

Common Pitfalls:
Students often misremember the counts of outcomes for each sum or assume that all sums from 2 to 12 are equally likely, which is not true. Another common error is to mistakenly include sum 7 when the condition requires strictly greater than 7. Using a table or diagram of the two dice grid helps maintain accuracy when counting.

Final Answer:
The probability that the sum of the two dice is greater than 7 is 5/12.