An oridinary cube has four faces, one face marked 2 another marked 3, Then the probability of obtaining a total of exactly 12 in five throws is

Options

A. `(5)/(1296)`
B. `(5)/(1944)`
C. `(5)/(2592)`
D. none of these

Answer

To solve the problem of finding the probability of obtaining a total of exactly 12 in five throws of a cube with faces marked 2 and 3, we can follow these steps:

### Step 1: Understand the Problem
We have a cube with four faces: one face marked 2, one face marked 3, and the other two faces are not specified (we will assume they are marked 0 for our calculations). We need to find the probability of getting a total of 12 from five throws of this cube.

### Step 2: Determine Possible Outcomes
In five throws, the maximum score we can achieve is 15 (if all throws yield a 3). To achieve a total of 12, we can have different combinations of 2s and 3s.

### Step 3: Identify Combinations
1. **Case 1:** Four throws yield 3 and one throw yields 0.
- Total = \(3 \times 4 + 0 = 12\)
- This means we have 4 occurrences of 3 and 1 occurrence of 0.

2. **Case 2:** Three throws yield 3 and two throws yield 2.
- Total = \(3 \times 3 + 2 \times 2 = 9 + 4 = 12\)
- This means we have 3 occurrences of 3 and 2 occurrences of 2.

### Step 4: Calculate the Probability for Each Case
**Case 1:**
- Probability of rolling a 3: \(P(3) = \frac{1}{6}\)
- Probability of rolling a 0 (assuming the other two faces are 0): \(P(0) = \frac{4}{6} = \frac{2}{3}\)

The probability for this case can be calculated using the binomial probability formula:
\[ P(\text{Case 1}) = \binom{5}{4} \left(\frac{1}{6}\right)^4 \left(\frac{2}{3}\right)^1 \]
Calculating this gives:
\[ P(\text{Case 1}) = 5 \cdot \left(\frac{1}{6}\right)^4 \cdot \left(\frac{2}{3}\right) = 5 \cdot \frac{1}{1296} \cdot \frac{2}{3} = \frac{10}{3888} \]

**Case 2:**
- Probability of rolling a 3: \(P(3) = \frac{1}{6}\)
- Probability of rolling a 2: \(P(2) = \frac{1}{6}\)

The probability for this case can be calculated similarly:
\[ P(\text{Case 2}) = \binom{5}{3} \left(\frac{1}{6}\right)^3 \left(\frac{1}{6}\right)^2 \]
Calculating this gives:
\[ P(\text{Case 2}) = 10 \cdot \left(\frac{1}{6}\right)^5 = 10 \cdot \frac{1}{7776} \]

### Step 5: Combine the Probabilities
Now, we add the probabilities from both cases:
\[ P(\text{Total}) = P(\text{Case 1}) + P(\text{Case 2}) = \frac{10}{3888} + \frac{10}{7776} \]

To add these fractions, we need a common denominator:
\[ P(\text{Total}) = \frac{20}{7776} + \frac{10}{7776} = \frac{30}{7776} \]

### Step 6: Simplify the Probability
Now we simplify:
\[ P(\text{Total}) = \frac{5}{1296} \]

### Final Answer
The probability of obtaining a total of exactly 12 in five throws is:
\[ \frac{5}{1296} \]

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