A fair die is thrown 20 times. The probability that on the 10th throw, the fourth six appears is `^20 C_(10)xx5^6//6^(20)` b. `120xx5^7//6^(10)` c. `84xx5^6//6^(10)` d. none of these

Options

A. `""^(20)C_(10)xx5//""^(6)//6""^(20)`
B. `120xx5^(7)//6^(10)`
C. `84xx5^(6)//6^(10)`
D. None of these

Explanation

To solve the problem, we need to find the probability that the fourth six appears on the 10th throw when a fair die is thrown 20 times.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need the fourth six to appear on the 10th throw. This means that in the first 9 throws, there must be exactly 3 sixes.

2. **Using Binomial Probability**:
The number of ways to get exactly 3 sixes in the first 9 throws can be calculated using the binomial coefficient \( \binom{n}{r} \), where \( n \) is the total number of trials (9 in this case) and \( r \) is the number of successes (3 sixes).

\[ \text{Number of ways} = \binom{9}{3} \]

3. **Calculating the Probability of Getting 3 Sixes**:
The probability of getting a six on a fair die is \( \frac{1}{6} \) and the probability of not getting a six is \( \frac{5}{6} \). Therefore, the probability of getting exactly 3 sixes in 9 throws is given by:

\[ P(X = 3) = \binom{9}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{6} \]

4. **Calculating the Probability of Getting the Fourth Six on the 10th Throw**:
The probability that the 10th throw is a six is simply \( \frac{1}{6} \).

5. **Combining the Probabilities**:
The total probability that the fourth six appears on the 10th throw is the product of the probability of getting exactly 3 sixes in the first 9 throws and the probability of getting a six on the 10th throw:

\[ P(\text{4th six on 10th throw}) = P(X = 3) \times P(\text{10th throw is a six}) \]

\[ P(\text{4th six on 10th throw}) = \binom{9}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{6} \times \frac{1}{6} \]

6. **Calculating the Final Expression**:
Substituting the values:

\[ P(\text{4th six on 10th throw}) = \binom{9}{3} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{6} \]

\[ = 84 \times \left(\frac{1}{6}\right)^4 \times \left(\frac{5}{6}\right)^{6} \]

\[ = 84 \times \frac{1}{6^4} \times \frac{5^6}{6^6} \]

\[ = 84 \times \frac{5^6}{6^{10}} \]

### Final Answer:
Thus, the probability that on the 10th throw, the fourth six appears is:

\[ \frac{84 \times 5^6}{6^{10}} \]

### Conclusion:
The correct option is (c) \( 84 \times \frac{5^6}{6^{10}} \).