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: \[ \frac{1}{8} \]

Answer 2

Step 1: Understanding the Problem
We are asked to find the probability of getting a total of 10 in a single throw of three fair dice.
Each die has 6 faces, so the total number of possible outcomes when throwing three dice is:
\[ 6 \times 6 \times 6 = 216 \]
This is the total number of possible outcomes. 
Step 2: Finding Favorable Outcomes
We need to count the number of favorable outcomes where the sum of the three dice is 10. This can be done by finding all possible combinations of numbers on the three dice that add up to 10.
The possible combinations are:
\[ (1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4), (3, 3, 4) \]
For each of these combinations, we can rearrange the dice in different ways. We can list all the permutations of these combinations:
- For \( (1, 3, 6) \), the permutations are: \( (1, 3, 6), (1, 6, 3), (3, 1, 6), (3, 6, 1), (6, 1, 3), (6, 3, 1) \) — 6 permutations.
- For \( (1, 4, 5) \), the permutations are: \( (1, 4, 5), (1, 5, 4), (4, 1, 5), (4, 5, 1), (5, 1, 4), (5, 4, 1) \) — 6 permutations.
- For \( (2, 2, 6) \), the permutations are: \( (2, 2, 6), (2, 6, 2), (6, 2, 2) \) — 3 permutations.
- For \( (2, 3, 5) \), the permutations are: \( (2, 3, 5), (2, 5, 3), (3, 2, 5), (3, 5, 2), (5, 2, 3), (5, 3, 2) \) — 6 permutations.
- For \( (2, 4, 4) \), the permutations are: \( (2, 4, 4), (4, 2, 4), (4, 4, 2) \) — 3 permutations.
- For \( (3, 3, 4) \), the permutations are: \( (3, 3, 4), (3, 4, 3), (4, 3, 3) \) — 3 permutations.
Now, we add up the total number of favorable outcomes:
\[ 6 + 6 + 3 + 6 + 3 + 3 = 27 \] Step 3: Calculating the Probability
The probability is the ratio of favorable outcomes to the total number of outcomes: \[ P(\text{sum} = 10) = \frac{27}{216} = \frac{1}{8} \] Step 4: Conclusion
Therefore, the probability of getting a sum of 10 in a single throw of three fair dice is \( \frac{1}{8} \).