Let S be the event that a vehicle goes straight, L be the event that a vehicle turns left, and R be the event that a vehicle turns right.

The probability that one vehicle goes straight and the other two turn left is given by:

\(P(SLL) = (0.3)(0.55)(0.55) = 0.09075\)

The probability that one vehicle goes straight and the other two turn right is given by:

\(P(SRR) = (0.3)(0.15)(0.15) = 0.00675\)

There are 3 ways to arrange SLL (SSL, SLL, LSL) and 3 ways to arrange SRR (SSR, SRR, RSR), so the total probability is:

\(\binom{3}{1} \times P(SLL) + \binom{3}{1} \times P(SRR) = 3 \times 0.09075 + 3 \times 0.00675 = 0.27225 + 0.02025 = 0.2925\)

Thus, the probability is approximately 0.293.