Step 1: Understanding the Concept:
We use the negative binomial distribution. For the third 5 to occur on the 7th throw, there must be exactly two 5's in the first 6 throws, and the 7th throw must be a 5.
Step 2: Detailed Explanation:
Probability of success (getting a 5) \( p = 1/6 \). Probability of failure (not getting a 5) \( q = 5/6 \).
Number of ways to get two 5's in 6 throws is \( \binom{6}{2} \).
\( P(\text{2 successes in 6 throws}) = \binom{6}{2} p^2 q^4 = 15 \times (1/6)^2 \times (5/6)^4 \).
The 7th throw must be a 5: \( P(\text{7th is 5}) = 1/6 \).
Total probability: \( 15 \times (1/6)^2 \times (5/6)^4 \times (1/6) = 15 \times \frac{625}{6^7} = \frac{15 \times 625}{279936} = \frac{9375}{279936} = \frac{3125}{93312} \).
Wait, checking calculations: \( \frac{15 \times 625}{279936} = \frac{9375}{279936} = \frac{3125}{93312} \). The provided option (D) is \( 6250/93312 \). Checking \( 2 \times 3125/93312 \).
Step 3: Final Answer:
Based on standard probability calculation, the result is \( 3125/93312 \).