The problem involves a binomial distribution where the probability of passing, \(p\), is 0.8, and the number of trials, \(n\), is 9. We need to find the probability that at most 6 students pass, which is \(P(X \leq 6)\).

This can be calculated as:

\(P(X \leq 6) = 1 - (P(7) + P(8) + P(9))\)

Using the binomial probability formula:

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

Calculate \(P(7)\), \(P(8)\), and \(P(9)\):

\(P(7) = \binom{9}{7} (0.8)^7 (0.2)^2\)

\(P(8) = \binom{9}{8} (0.8)^8 (0.2)^1\)

\(P(9) = \binom{9}{9} (0.8)^9 (0.2)^0\)

Substitute these into the equation:

\(P(X \leq 6) = 1 - (0.3019899 + 0.3019899 + 0.1342177)\)

\(P(X \leq 6) = 1 - 0.7381975 = 0.2618025\)

Rounding to three decimal places, the probability is \(0.262\).