Question

Three fair dice are thrown simultaneously. What is the probability that the sum of numbers on their tops is at least \(6\)?

• A. \(\frac{5}{108}\)
• B. \(\frac{1}{24}\)
• C. \(\frac{103}{108}\)
• D. \(\frac{17}{18}\)

Hint:

Using the complementary probability method often simplifies counting in dice problems, especially when the condition is "at least" or "at most".

Answer 1

The Correct Option is C

Step 1: Total possible outcomes.
Each die has 6 faces.
Number of total outcomes when three dice are thrown: \[ 6 \times 6 \times 6 = 216 \]
Step 2: Complementary probability approach.
Instead of counting outcomes with sum \(\geq 6\), we calculate probability of sum \(< 6\) and subtract from 1.

Step 3: Possible sums less than 6.
Smallest possible sum is \(3\) (i.e., \(1+1+1\)).
We need sums: \(3, 4, 5\).

Step 4: Counting outcomes.
- Sum = 3: Only (1,1,1) → \(1\) outcome.
- Sum = 4: Possible combinations: (1,1,2) and permutations.
Number of permutations = \(\frac{3!}{2!} = 3\) outcomes when the "2" appears in one place. Also (1,2,1) and (2,1,1) are included already in permutations, so total \(3\) outcomes. No other combination because numbers can't exceed 6.
- Sum = 5: Combinations: (1,1,3), (1,3,1), (3,1,1), and (1,2,2), (2,1,2), (2,2,1).
For (1,1,3): permutations = \(\frac{3!}{2!} = 3\).
For (1,2,2): permutations = \(\frac{3!}{2!} = 3\).
Total for sum = 5 → \(3 + 3 = 6\) outcomes.

Step 5: Total outcomes with sum \(< 6\).
\[ \text{Count} = 1 + 3 + 6 = 10 \]
Step 6: Probability.
\[ P(\text{sum} \geq 6) = 1 - \frac{10}{216} = \frac{216 - 10}{216} = \frac{206}{216} = \frac{103}{108} \] \[ \boxed{\frac{103}{108}} \]