Question

The probability of getting 10 in a single throw of three fair dice is:

• A. \( \frac{1}{6} \)
• B. \( \frac{1}{8} \)
• C. \( \frac{1}{9} \)
• D. \( \frac{1}{5} \)

Hint:

To find the probability of a specific outcome in a dice game, first determine all the possible outcomes, then count the favorable outcomes.

Answer 1

The Correct Option is B

The total number of possible outcomes when rolling three dice is calculated as \(6 \times 6 \times 6 = 216\). To determine the number of outcomes where the sum of the three dice is 10, all possible combinations summing to 10 are enumerated. The analysis reveals 27 such favorable outcomes. The outcomes are categorized as follows:

• Case 1: Combinations with numbers \(1, 3, 6\) yield \( \frac{3!}{1!} = 6 \) outcomes.
• Case 2: Combinations with numbers \(1, 4, 5\) yield \( \frac{3!}{1!} = 6 \) outcomes.
• Case 3: Combinations with numbers \(2, 2, 6\) yield \( \frac{3!}{2!} = 3 \) outcomes.
• Case 4: Combinations with numbers \(2, 3, 5\) yield \( \frac{3!}{1!} = 6 \) outcomes.
• Case 5: Combinations with numbers \(2, 4, 4\) yield \( \frac{3!}{2!} = 3 \) outcomes.
• Case 6: Combinations with numbers \(3, 3, 4\) yield \( \frac{3!}{2!} = 3 \) outcomes.

The sum of these favorable outcomes is:\[\text{Favourable outcomes} = 27\]The probability of achieving a sum of 10 is therefore:\[\text{Probability} = \frac{27}{216} = \frac{1}{8}\]This problem leads to the quadratic equation:\[(x + 24)(x - 20) = 0\]The solution for \( x \) is:\[x = 20\]Consequently, the probability is:\[\frac{1}{8}\]