Let the random variable \(X\) represent the number of buses in the sample of 11 vehicles. Since 16% of the vehicles are buses, the probability of selecting a bus is \(p = 0.16\).

We need to find \(P(X < 3) = P(0) + P(1) + P(2)\).

Using the binomial probability formula \(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\), where \(n = 11\) and \(p = 0.16\):

\(P(0) = \binom{11}{0} (0.16)^0 (0.84)^{11} = (0.84)^{11} = 0.1469\)

\(P(1) = \binom{11}{1} (0.16)^1 (0.84)^{10} = 11 \times 0.16 \times (0.84)^{10} = 0.30782\)

\(P(2) = \binom{11}{2} (0.16)^2 (0.84)^9 = 55 \times (0.16)^2 \times (0.84)^9 = 0.2931\)

Therefore, \(P(X < 3) = 0.1469 + 0.30782 + 0.2931 = 0.748\).