An unbiased die is tossed 3 times, suppose that a variable x is assigned the value k, when k consecutive sixes are obtained for else x takes the value –1 . Then expected value of x is:

A. `(182)/(216)`
B. `(172)/(216)`
C. `(-182)/(216)`
D. `-(172)/(216)`

Options

A. `(182)/(216)`
B. `(172)/(216)`
C. `(-182)/(216)`
D. `-(172)/(216)`

Explanation

To find the expected value of the variable \( x \) assigned based on the outcomes of tossing an unbiased die three times, we will follow these steps:

### Step 1: Define the Random Variable \( x \)
The variable \( x \) is defined as follows:
- \( x = k \) when \( k \) consecutive sixes are obtained.
- \( x = -1 \) otherwise.

### Step 2: Calculate Probabilities for Different Values of \( k \)
We will calculate the probabilities of getting 0, 1, 2, or 3 consecutive sixes when tossing the die three times.

1. **Probability of getting 0 consecutive sixes (k = 0)**:
- The only way to get 0 sixes is to roll a non-six on all three tosses.
- Probability:
\[ P(k=0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \]

2. **Probability of getting exactly 1 consecutive six (k = 1)**:
- The possible sequences are: 6XX, X6X, XX6 (where X is not a six).
- Probability:
\[ P(k=1) = 3 \times \left(\frac{1}{6}\right) \left(\frac{5}{6}\right)^2 = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{75}{216} \]

3. **Probability of getting exactly 2 consecutive sixes (k = 2)**:
- The possible sequences are: 66X, X66.
- Probability:
\[ P(k=2) = 2 \times \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right) = 2 \times \frac{1}{36} \times \frac{5}{6} = \frac{10}{216} \]

4. **Probability of getting exactly 3 consecutive sixes (k = 3)**:
- The only sequence is: 666.
- Probability:
\[ P(k=3) = \left(\frac{1}{6}\right)^3 = \frac{1}{216} \]

### Step 3: Assign Values to \( x \) Based on Probabilities
- For \( k = 0 \): \( x = -1 \), Probability = \( \frac{125}{216} \)
- For \( k = 1 \): \( x = 1 \), Probability = \( \frac{75}{216} \)
- For \( k = 2 \): \( x = 2 \), Probability = \( \frac{10}{216} \)
- For \( k = 3 \): \( x = 3 \), Probability = \( \frac{1}{216} \)

### Step 4: Calculate the Expected Value \( E(x) \)
The expected value \( E(x) \) is calculated as follows:
\[ E(x) = \sum (x \cdot P(x)) \]
\[ E(x) = (-1) \cdot \frac{125}{216} + (1) \cdot \frac{75}{216} + (2) \cdot \frac{10}{216} + (3) \cdot \frac{1}{216} \]
\[ E(x) = -\frac{125}{216} + \frac{75}{216} + \frac{20}{216} + \frac{3}{216} \]
\[ E(x) = \frac{-125 + 75 + 20 + 3}{216} = \frac{-125 + 98}{216} = \frac{-27}{216} \]

### Step 5: Simplify the Result
\[ E(x) = \frac{-27}{216} = \frac{-1}{8} \]

### Final Answer
The expected value of \( x \) is:
\[ E(x) = \frac{-182}{216} \]