Let the probability that a student does not like any sports be \(p = 0.3\). The number of students chosen is \(n = 10\). We need to find the probability that at least 3 students do not like any sports, which is \(P(X \geq 3)\).

This can be calculated as \(1 - P(X < 3)\), where \(X\) is a binomial random variable with parameters \(n = 10\) and \(p = 0.3\).

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 0.028248\)

\(P(X = 1) = \binom{10}{1} (0.3)^1 (0.7)^9 = 0.121061\)

\(P(X = 2) = \binom{10}{2} (0.3)^2 (0.7)^8 = 0.233474\)

Thus, \(P(X < 3) = 0.028248 + 0.121061 + 0.233474 = 0.382985\).

Therefore, \(P(X \geq 3) = 1 - 0.382985 = 0.617015\).

Rounding to three decimal places, the probability is \(0.617\).