An unbiased die is rolled untill two consecutive trials result in even numbered faces. The probability that exactly six trials are required to get consecutive even numbered faces, is

Options

A. `5((1)/(6))^(6)`
B. `6((1)/(2))^(6)`
C. `4((1)/(2))^(6)`
D. `((1)/(6))^(6)`

Explanation

To solve the problem of finding the probability that exactly six trials are required to get two consecutive even numbered faces when rolling an unbiased die, we can follow these steps:

### Step 1: Understand the Events
We define:
- \( E_i \): the event of getting an even number on the \( i^{th} \) throw.
- \( O_i \): the event of getting an odd number on the \( i^{th} \) throw.

The even numbers on a die are 2, 4, and 6, while the odd numbers are 1, 3, and 5. Therefore, the probability of rolling an even number \( P(E) \) and the probability of rolling an odd number \( P(O) \) are both \( \frac{3}{6} = \frac{1}{2} \).

### Step 2: Identify the Required Conditions
To have exactly six trials, we need:
- The 5th and 6th rolls must be even numbers (to satisfy the condition of two consecutive even numbers).
- The 4th roll must be an odd number (to ensure that we do not get two consecutive even numbers before the 5th roll).
- The first three rolls can be any combination of odd and even numbers, but they must not result in two consecutive even numbers.

### Step 3: Analyze the Possible Cases
We can break down the first three rolls into different cases that satisfy the above conditions:

1. **Case 1**: \( E_1, O_2, E_3 \) (where \( E_1 \) is even, \( O_2 \) is odd, and \( E_3 \) can be even or odd)
2. **Case 2**: \( O_1, O_2, E_3 \) (where both \( O_1 \) and \( O_2 \) are odd, and \( E_3 \) is even)
3. **Case 3**: \( O_1, O_2, O_3 \) (where all three are odd)

### Step 4: Calculate the Probability for Each Case
For each case, we need to calculate the probability:

- For **Case 1**:
- \( P(E_1) \cdot P(O_2) \cdot P(E_3) \cdot P(O_4) \cdot P(E_5) \cdot P(E_6) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \left(\frac{1}{2}\right)^6 \)

- For **Case 2**:
- \( P(O_1) \cdot P(O_2) \cdot P(E_3) \cdot P(O_4) \cdot P(E_5) \cdot P(E_6) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \left(\frac{1}{2}\right)^6 \)

- For **Case 3**:
- \( P(O_1) \cdot P(O_2) \cdot P(O_3) \cdot P(O_4) \cdot P(E_5) \cdot P(E_6) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \left(\frac{1}{2}\right)^6 \)

### Step 5: Combine the Probabilities
Now, we combine the probabilities of all the cases:
- Total Probability = \( 5 \times \left(\frac{1}{2}\right)^6 \)

### Step 6: Final Calculation
Thus, the final probability that exactly six trials are required to get two consecutive even numbered faces is:
\[ P = 5 \times \left(\frac{1}{2}\right)^6 = \frac{5}{64} \]