To solve this, we can use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n = 10$ (the number of trials, in this case, 10 rolls of the die),
• $k = 6$ (the number of times we want to roll a five),
• $p = \frac{1}{6}$ (the probability of rolling a five on any single roll),
• $1-p = \frac{5}{6}$ (the probability of not rolling a five).

Now, calculate the binomial coefficient $\binom{10}{6}$:

$\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Then, calculate the probability:

$P(X = 6) = 210 \times \left(\frac{1}{6}\right)^6 \times \left(\frac{5}{6}\right)^4$

Simplifying the powers:

$P(X = 6) = 210 \times \left(\frac{1}{7776}\right) \times \left(\frac{625}{1296}\right)$

Now calculate the result:

$P(X = 6) \approx 210 \times 0.0001286 \times 0.48225 \approx 0.013$

Thus, the correct answer is:

$\boxed{0.013}$

Hence, the correct choice is C (0.013).

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Would you like further details or have any questions?

Here are five related questions for deeper understanding:

1. What is the general form of the binomial probability distribution?
2. How do you calculate the binomial coefficient $\binom{n}{k}$?
3. What does the binomial distribution model, and when can it be used?
4. How would the probability change if the number of rolls increased?
5. What would be the probability of rolling at least one five in 10 rolls?

Tip: The binomial distribution is useful when each trial is independent, and there are only two possible outcomes per trial (success or failure).