An ordinary cube has 4 blank faces, one face mark 2 and another marked 3, then the probability of obtaining 12 in 5 throws is

Explanation

To find the probability of obtaining a sum of 12 in 5 throws of a cube with 4 blank faces, one face marked 2, and another face marked 3, we can break down the problem into manageable steps.

### Step-by-Step Solution:

1. **Understanding the Cube**:
The cube has 6 faces: 4 blank faces, 1 face marked with 2, and 1 face marked with 3.

2. **Possible Outcomes**:
To achieve a total of 12 in 5 throws, we can only achieve this with the following combinations:
- 4 throws of 3 and 1 throw of blank (3+3+3+3+0 = 12)
- 3 throws of 2 and 2 throws of 3 (2+2+2+3+3 = 12)

3. **Calculating the Probability for Each Case**:

**Case 1**: 4 Threes and 1 Blank
- Probability of rolling a 3: \( P(3) = \frac{1}{6} \)
- Probability of rolling a blank: \( P(\text{blank}) = \frac{4}{6} = \frac{2}{3} \)
- The probability of getting 4 threes and 1 blank in any order:
\[ P(\text{4 threes, 1 blank}) = \left( \frac{1}{6} \right)^4 \times \left( \frac{2}{3} \right)^1 \times \binom{5}{1} \]
\[ = \left( \frac{1}{6} \right)^4 \times \left( \frac{2}{3} \right) \times 5 \]

**Case 2**: 3 Twos and 2 Threes
- Probability of rolling a 2: \( P(2) = \frac{1}{6} \)
- The probability of getting 3 twos and 2 threes in any order:
\[ P(\text{3 twos, 2 threes}) = \left( \frac{1}{6} \right)^3 \times \left( \frac{1}{6} \right)^2 \times \binom{5}{3} \]
\[ = \left( \frac{1}{6} \right)^5 \times 10 \]

4. **Combining the Probabilities**:
Now, we combine the probabilities from both cases:
\[ P(\text{total}) = P(\text{4 threes, 1 blank}) + P(\text{3 twos, 2 threes}) \]
\[ = \left( \frac{1}{6} \right)^4 \times \left( \frac{2}{3} \right) \times 5 + \left( \frac{1}{6} \right)^5 \times 10 \]
\[ = \frac{5 \times 2}{3 \times 6^4} + \frac{10}{6^5} \]
\[ = \frac{10}{3 \times 6^4} + \frac{10}{6^5} \]

5. **Finding a Common Denominator**:
The common denominator is \( 3 \times 6^5 \):
\[ = \frac{10 \times 6}{3 \times 6^5} + \frac{10}{3 \times 6^5} \]
\[ = \frac{60 + 10}{3 \times 6^5} = \frac{70}{3 \times 6^5} \]

6. **Final Calculation**:
\[ = \frac{70}{3 \times 7776} = \frac{70}{23328} \]

7. **Simplifying**:
\[ = \frac{35}{11664} \]

Thus, the final probability of obtaining a sum of 12 in 5 throws is \( \frac{35}{11664} \).