Let the events be defined as follows:

• ext{SSS}: All three vehicles go straight on.
• ext{LLL}: All three vehicles turn left.
• ext{RRR}: All three vehicles turn right.

The probability that all three vehicles choose the same direction is given by:

\(P( ext{same direction}) = P( ext{SSS}) + P( ext{LLL}) + P( ext{RRR})\)

Calculating each probability:

\(P( ext{SSS}) = 0.3^3\)

\(P( ext{LLL}) = 0.55^3\)

\(P( ext{RRR}) = 0.15^3\)

Thus,

\(P( ext{same direction}) = 0.3^3 + 0.55^3 + 0.15^3\)

We need to find the probability that all three vehicles go straight on given that they all choose the same direction:

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{P( ext{SSS and same direction})}{P( ext{same direction})}\)

Since ext{SSS} is a subset of ext{same direction},

\(P( ext{SSS and same direction}) = P( ext{SSS}) = 0.3^3\)

Therefore,

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{0.3^3}{0.3^3 + 0.55^3 + 0.15^3}\)

Calculating the values:

\(0.3^3 = 0.027\)

\(0.55^3 = 0.166375\)

\(0.15^3 = 0.003375\)

Thus,

\(P( ext{same direction}) = 0.027 + 0.166375 + 0.003375 = 0.197\)

Finally,

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{0.027}{0.197} \approx 0.137\)