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

Total outcomes when rolling three dice is \(6 \times 6 \times 6 = 216\). To calculate the number of cases where the sum is 10, we list all combinations of numbers on the dice that add to 10. After listing the cases, we find there are 27 favorable outcomes. We consider different cases of outcomes:

• Case 1: \(1 + 3 + 6 \rightarrow \text{outcomes} = \frac{3!}{1!} = 6\)
• Case 2: \(1 + 4 + 5 \rightarrow \text{outcomes} = \frac{3!}{1!} = 6\)
• Case 3: \(2 + 2 + 6 \rightarrow \text{outcomes} = \frac{3!}{2!} = 3\)
• Case 4: \(2 + 3 + 5 \rightarrow \text{outcomes} = \frac{3!}{1!} = 6\)
• Case 5: \(2 + 4 + 4 \rightarrow \text{outcomes} = \frac{3!}{2!} = 3\)
• Case 6: \(3 + 3 + 4 \rightarrow \text{outcomes} = \frac{3!}{2!} = 3\)

Sum of favorable outcomes: \[ \text{Favourable outcomes} = 27 \] Thus, the probability is: \[ \text{Probability} = \frac{27}{216} = \frac{1}{8} \] This leads to the quadratic equation: \[ (x + 24)(x - 20) = 0 \] Solving for \( x \), we find: \[ x = 20 \] Thus, the probability is: \[ \frac{1}{8} \]