An ordinary cube has four blank faces, one face marked 2 and another marked 3. Then the probability of obtaining 9 in 5 throws, is

Options

A. `(31)/(7776)`
B. `(5)/(2592)`
C. `(5)/(1944)`
D. `(5)/(1296)`

Answer

To solve the problem, we need to find the probability of obtaining a total of 9 when throwing a specially marked cube 5 times. The cube has 4 blank faces, one face marked with 2, and another face marked with 3.

### Step-by-Step Solution:

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

2. **Total Outcomes**:
- When throwing the cube 5 times, the total number of possible outcomes is given by \(6^5\) since there are 6 faces on the cube.
- Calculation:
\[ 6^5 = 7776 \]

3. **Finding Favorable Outcomes**:
- We need to find combinations of the numbers that add up to 9 in 5 throws. The possible combinations of faces that can achieve this are:
- 0 times 3 and 4 times 2: \(0 \times 3 + 4 \times 2 = 8\) (not valid)
- 1 time 3 and 3 times 2: \(1 \times 3 + 3 \times 2 = 9\) (valid)
- 2 times 3 and 1 time 2: \(2 \times 3 + 1 \times 2 = 8\) (not valid)
- 3 times 3 and 0 times 2: \(3 \times 3 + 0 \times 2 = 9\) (valid)

4. **Calculating Combinations**:
- For the combination of 1 time 3 and 3 times 2:
- The number of ways to choose which throw is 3 (out of 5) is given by \( \binom{5}{1} \).
- The remaining 4 throws must be 2s, so the number of ways to arrange this is:
\[ \binom{5}{1} = 5 \]

- For the combination of 3 times 3 and 2 times blank:
- The number of ways to choose which 3 throws are 3s (out of 5) is given by \( \binom{5}{3} \).
- The remaining 2 throws must be blanks, so the number of ways to arrange this is:
\[ \binom{5}{3} = 10 \]

5. **Total Favorable Outcomes**:
- Adding the favorable outcomes from both combinations:
\[ 5 + 10 = 15 \]

6. **Calculating Probability**:
- The probability of obtaining a total of 9 in 5 throws is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{15}{7776} \]

7. **Simplifying the Probability**:
- Simplifying \( \frac{15}{7776} \):
\[ P = \frac{5}{2592} \]

### Final Answer:
The probability of obtaining a total of 9 in 5 throws is \( \frac{5}{2592} \).