I don’t agree:

This is a binomial probability problem. Each roll of a fair six-sided die is an independent trial with success probability \(p = \frac{1}{6}\) (rolling a 6) and failure probability \(1 - p = \frac{5}{6}\) (rolling anything else). We want the probability of exactly \(k = 5\) successes in \(n = 8\) trials.

The binomial probability formula is:

\[

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

\]

First, compute the binomial coefficient:

\[

\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5! \cdot 3!} = 56

\]

Substitute into the formula:

\[

P(X = 5) = 56 \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^3 = 56 \cdot \frac{1^5 \cdot 5^3}{6^8} = 56 \cdot \frac{125}{1,679,616} = \frac{7,000}{1,679,616}

\]

Simplify the fraction by dividing numerator and denominator by their greatest common divisor (which is 8):

\[

\frac{7,000 \div 8}{1,679,616 \div 8} = \frac{875}{209,952}

\]

This fraction is in lowest terms. As a decimal approximation, this is about 0.00417 (or 0.417%).