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

Concept: The phrase "at least 10" in probability means we are looking for outcomes where the sum (\(S\)) of the two face values satisfies the inequality: \[ S \ge 10 \quad \implies \quad S = 10, \text{ or } S = 11, \text{ or } S = 12 \] The standard probability formula is: \[ P = \frac{\text{Number of Favorable Outcomes } (n(E))}{\text{Total Number of Sample Outcomes } (n(S))} \]

Step 1: Determining the total sample space size.
When rolling a single 6-sided die, there are 6 unique outcomes. For a pair of dice thrown simultaneously, the total number of sample outcomes in the grid space is: \[ n(S) = 6 \times 6 = 36 \]

Step 2: Listing the favorable coordinates matching our condition.
Let's list out all possible coordinate pairs \((d_1, d_2)\) that yield our targeted sum thresholds:
• For Sum = 10: \((4, 6), (5, 5), (6, 4)\) \(\rightarrow\) 3 ways
• For Sum = 11: \((5, 6), (6, 5)\) \(\rightarrow\) 2 ways
• For Sum = 12: \((6, 6)\) \(\rightarrow\) 1 way Summing these discrete valid configurations together: \[ n(E) = 3 + 2 + 1 = 6 \text{ favorable ways} \]

Step 3: Evaluating the final probability ratio.
\[ P(S \ge 10) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6} \]