A fair die is tossed eight times. The probability that a third six is observed in eight throw is `(\ ^7C_(10)xx5^7)/(6^7)` b. `(\ ^7C_2xx5^2)/(6^8)` c. `(\ ^7C_2xx5^5)/(6^6)` d. none of these

Options

A. `(.^7C_(2)xx5^5)/(6^7)`
B. `(.^7C_(2)xx5^5)/(6^8)`
C. `(.^7C_(2)xx5^5)/(6^6)`
D. none of these

Explanation

To solve the problem of finding the probability that a third six is observed on the eighth throw when a fair die is tossed eight times, we can follow these steps:

### Step 1: Understand the Problem
We need to find the probability that the third occurrence of rolling a six happens specifically on the eighth toss. This means that in the first seven tosses, we must have exactly two sixes.

**Hint**: Identify the conditions that must be satisfied for the event to occur.

### Step 2: Determine the Number of Trials
Since we are interested in the first seven tosses, we denote this as \( n = 7 \). We need to find the number of ways to get exactly 2 sixes in these 7 tosses.

**Hint**: Use combinations to count the number of ways to choose which of the 7 tosses will be sixes.

### Step 3: Calculate the Combinations
The number of ways to choose 2 sixes from 7 tosses is given by the binomial coefficient \( \binom{7}{2} \).

**Hint**: Recall the formula for combinations: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).

### Step 4: Calculate the Probabilities
The probability of rolling a six on a single toss is \( \frac{1}{6} \), and the probability of not rolling a six (rolling a 1, 2, 3, 4, or 5) is \( \frac{5}{6} \).

For our scenario:
- We need 2 sixes: \( \left(\frac{1}{6}\right)^2 \)
- We need 5 non-sixes: \( \left(\frac{5}{6}\right)^5 \)

**Hint**: Remember to multiply the probabilities of the successes and failures.

### Step 5: Combine the Results
The total probability of getting exactly 2 sixes in the first 7 tosses followed by a six on the eighth toss is given by:
\[ P = \binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \times \left(\frac{1}{6}\right) \]
This simplifies to:
\[ P = \binom{7}{2} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^5 \]

### Step 6: Final Calculation
Now, substituting the values:
\[ P = \binom{7}{2} \cdot \frac{1}{6^3} \cdot \frac{5^5}{6^5} = \binom{7}{2} \cdot \frac{5^5}{6^8} \]

### Step 7: Identify the Correct Option
Now we can compare our result with the given options. The expression we derived matches option (b):
\[ \frac{\binom{7}{2} \cdot 5^5}{6^8} \]

### Conclusion
Thus, the probability that a third six is observed on the eighth throw is given by option (b).

### Final Answer
The answer is \( \frac{\binom{7}{2} \cdot 5^5}{6^8} \).

---