Question

A pair of fair dice is thrown simultaneously. What is the probability that the sum of the numbers appearing on the top faces is at least 10?

• A. \( \frac{1}{6} \)
• B. \( \frac{1}{12} \)
• C. \( \frac{5}{36} \)
• D. \( \frac{1}{4} \)

Hint:

For dice-sum questions, memorize the Symmetric Triangular Rule to count outcomes instantly without writing them out: The number of ways to get a sum follows a perfect hill shape that peaks at a sum of 7 (6 ways). For sums above 7, the number of ways is simply \(14 - \text{Sum}\).
• Ways to get 10 = \(14 - 10 = 4\) Adding them up directly: \(3 + 2 + 1 = 6\) ways, saving you valuable time!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the probability that the sum of the numbers on the top faces of two fair dice thrown simultaneously is "at least 10" (i.e., the sum \(S\) satisfies \(S \ge 10\)).
Step 2: Key Formula or Approach:
The probability of an event \(E\) is:
\[ P(E) = \frac{\text{Number of Favorable Outcomes } (n(E))}{\text{Total Number of Sample Outcomes } (n(S))} \] For a pair of six-sided dice, the total number of sample outcomes is:
\[ n(S) = 6 \times 6 = 36 \] The event "at least 10" includes the sums of 10, 11, and 12.
Step 3: Detailed Explanation:
1. List the favorable coordinate pairs \((d_1, d_2)\) for each possible sum \(S \ge 10\):
- For Sum = 10: \((4, 6), (5, 5), (6, 4)\) \(\rightarrow\) 3 outcomes
- For Sum = 11: \((5, 6), (6, 5)\) \(\rightarrow\) 2 outcomes
- For Sum = 12: \((6, 6)\) \(\rightarrow\) 1 outcome
2. Sum the individual counts to find the total number of favorable outcomes:
\[ n(E) = 3 + 2 + 1 = 6 \text{ outcomes} \] 3. Calculate the probability:
\[ P(S \ge 10) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6} \] This matches Option (A).
Step 4: Final Answer:
The probability that the sum is at least 10 is \( \frac{1}{6} \), which corresponds to Option (A).