A fair die is thrown again and again until three sixes are obtained . The probability of obtaining third six in the eighth thrown is

Options

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

Answer

To solve the problem of finding the probability of obtaining the third six on the eighth throw of a fair die, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the probability that the third six occurs on the eighth throw of a fair die. This means that in the first seven throws, we must have exactly two sixes, and then we must roll a six on the eighth throw.

2. **Define the Probabilities**:
The probability of rolling a six (success) on a fair die is \( P(S) = \frac{1}{6} \).
The probability of not rolling a six (failure) is \( P(F) = \frac{5}{6} \).

3. **Determine the Required Outcomes**:
For the third six to occur on the eighth throw, we need:
- Exactly 2 sixes in the first 7 throws.
- A six on the eighth throw.

4. **Calculate the Combinations**:
The number of ways to choose 2 throws from the first 7 throws where we will get sixes is given by the combination formula \( \binom{n}{k} \), where \( n \) is the total number of trials (7) and \( k \) is the number of successes (2):
\[ \text{Number of ways} = \binom{7}{2} \]

5. **Calculate the Probability of the Outcomes**:
The probability of getting exactly 2 sixes in 7 throws and 5 non-sixes can be calculated as follows:
\[ P(\text{2 sixes in 7 throws}) = \binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \]

6. **Include the Probability of the Eighth Throw**:
Since we need a six on the eighth throw, we multiply the previous probability by \( \frac{1}{6} \):
\[ P(\text{third six on 8th throw}) = \binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \cdot \frac{1}{6} \]

7. **Final Calculation**:
Putting it all together:
\[ P(\text{third six on 8th throw}) = \binom{7}{2} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^5 \]
\[ = \frac{7 \times 6}{2 \times 1} \cdot \left(\frac{1}{6}\right)^3 \cdot \left(\frac{5}{6}\right)^5 \]
\[ = 21 \cdot \frac{1}{216} \cdot \frac{3125}{7776} \]
\[ = \frac{21 \cdot 3125}{16632} \]
\[ = \frac{65625}{16632} \]
(You can simplify this fraction further if needed.)

### Final Answer:
The probability of obtaining the third six on the eighth throw is \( \frac{21 \cdot 3125}{7776} \) or approximately \( 0.0635 \).