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. `3//177`
B. `5//259`
C. `5//194`
D. `5//162`

Answer

To solve the problem of finding the probability of obtaining a total of 9 in 5 throws of a cube with 4 blank faces, one face marked 2, and one face marked 3, we can break the solution down into clear steps.

### Step-by-Step Solution:

1. **Identify Possible Outcomes**:
The cube has the following faces:
- 4 blank faces (0)
- 1 face marked 2
- 1 face marked 3

Therefore, the possible outcomes when rolling the cube are 0, 0, 0, 0, 2, and 3.

2. **Determine Combinations to Achieve a Total of 9**:
We need to find combinations of the numbers that can sum up to 9 in 5 throws. The two possible cases are:
- Case 1: Obtaining three 2s and one 3, with one blank face.
- Case 2: Obtaining three 3s and two blank faces.

3. **Calculate the Number of Ways for Each Case**:
- **Case 1**: Three 2s and one 3
- We need to select 3 positions out of 5 for the 2s, and then select 1 position out of the remaining 2 for the 3.
- The number of ways to choose 3 positions from 5 is given by \( \binom{5}{3} \).
- The number of ways to choose 1 position from the remaining 2 is given by \( \binom{2}{1} \).
- Thus, the total ways for Case 1 = \( \binom{5}{3} \times \binom{2}{1} \).

- **Case 2**: Three 3s
- Here, we need to select 3 positions out of 5 for the 3s.
- The number of ways to choose 3 positions from 5 is \( \binom{5}{3} \).
- Thus, the total ways for Case 2 = \( \binom{5}{3} \).

4. **Calculate Total Ways**:
- The total number of ways to achieve a total of 9 in 5 throws is the sum of the ways from both cases:
\[ \text{Total Ways} = \left( \binom{5}{3} \times \binom{2}{1} \right) + \left( \binom{5}{3} \right) \]

5. **Calculate the Total Number of Outcomes**:
- The total number of outcomes when rolling the cube 5 times is \( 6^5 \) because there are 6 faces on the cube.

6. **Calculate the Probability**:
- The probability of obtaining a total of 9 in 5 throws is given by:
\[ P = \frac{\text{Total Ways}}{6^5} \]

7. **Final Calculation**:
- Calculate \( \binom{5}{3} = 10 \) and \( \binom{2}{1} = 2 \).
- Therefore, Total Ways = \( (10 \times 2) + 10 = 20 + 10 = 30 \).
- The total number of outcomes = \( 6^5 = 7776 \).
- Thus, the probability \( P = \frac{30}{7776} = \frac{5}{1296} \).

### Final Answer:
The probability of obtaining a total of 9 in 5 throws is \( \frac{5}{1296} \).