In a random experiment, a fair die is rolled until two fours are obtained in succession.The probability that the experiment will end in the fifth throw of the die is equal to

Options

A. `(150)/(6^(5))`
B. `(175)/(6^(5))`
C. `(200)/(6^(5))`
D. `(225)/(6^(5))`

Answer

To find the probability that the experiment will end in the fifth throw of the die when rolling a fair die until two consecutive fours are obtained, we can break down the problem as follows:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to determine the conditions under which the experiment ends on the fifth throw. This means that the fifth throw must be a four, and the fourth throw must also be a four. The first three throws must not result in two consecutive fours.

2. **Possible Outcomes**:
The only way for the experiment to end on the fifth throw is if the outcomes of the first four throws are such that:
- The fourth throw is a four (4).
- The fifth throw is also a four (4).
- The first three throws must not contain two consecutive fours.

3. **Identifying Valid Sequences**:
The valid sequences for the first three throws can be:
- Any combination of numbers from {1, 2, 3, 5, 6} (not a four) and one four (but not consecutive).
- The valid patterns for the first three throws can be:
- Non-four, Non-four, Non-four (NNN)
- Non-four, Non-four, Four (NN4)
- Non-four, Four, Non-four (N4N)
- Four, Non-four, Non-four (4NN)

4. **Calculating Probabilities**:
- The probability of rolling a non-four (1, 2, 3, 5, or 6) is \( \frac{5}{6} \).
- The probability of rolling a four is \( \frac{1}{6} \).

5. **Calculating the Probability for Each Pattern**:
- For the first three throws being non-fours:
- Probability = \( \left( \frac{5}{6} \right)^3 \)
- For the fourth and fifth throws being fours:
- Probability = \( \left( \frac{1}{6} \right)^2 \)

6. **Combining the Probabilities**:
The total probability that the experiment ends on the fifth throw can be calculated as:
\[ P(\text{ending on 5th throw}) = P(\text{first three throws}) \times P(\text{fourth throw}) \times P(\text{fifth throw}) \]
\[ = \left( \frac{5}{6} \right)^3 \times \left( \frac{1}{6} \right) \times \left( \frac{1}{6} \right) \]

7. **Calculating the Final Probability**:
\[ = \left( \frac{5}{6} \right)^3 \times \left( \frac{1}{6} \right)^2 \]
\[ = \frac{125}{216} \times \frac{1}{36} \]
\[ = \frac{125}{1296} \]

### Final Answer:
The probability that the experiment will end in the fifth throw of the die is \( \frac{125}{1296} \).