Let the probability that a student plays at least one musical instrument be 0.72 (since 28% do not play any instrument).

We need to find the probability that more than 9 students play at least one instrument in a sample of 12 students. This is equivalent to finding:

\(P(X > 9) = P(10) + P(11) + P(12)\)

Using the binomial probability formula:

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

where \(n = 12\), \(p = 0.72\), and \(1-p = 0.28\).

Calculate each probability:

\(P(10) = \binom{12}{10} (0.72)^{10} (0.28)^2\)

\(P(11) = \binom{12}{11} (0.72)^{11} (0.28)^1\)

\(P(12) = \binom{12}{12} (0.72)^{12} (0.28)^0\)

Substitute and calculate:

\(P(10) = 66 \times 0.1934917632 \times 0.0784 = 0.193725\)

\(P(11) = 12 \times 0.1393140695 \times 0.28 = 0.0905726\)

\(P(12) = 1 \times 0.1934917632 \times 1 = 0.0190484\)

Sum these probabilities:

\(P(X > 9) = 0.193725 + 0.0905726 + 0.0190484 = 0.303346\)

Thus, the probability that more than 9 students play at least one musical instrument is approximately 0.304.