The first thing to notice is that the event of seeing a car is a Poisson process. Recall that in the Poisson distribution, the probability of $k$ events occurring within a timeframe $t$ is $$ P(N=k) = \frac{ (\lambda t)^k e^{- \lambda t}}{k!} $$ where $\lambda$ is the mean event rate. Now, let's consider the probability of seeing no cars in the time window. This probability is $$ P(N=0) = \frac{(\lambda t)^0 e^{- \lambda t}}{0!} = e^{- \lambda t}. $$ We can use this formula to derive a relationship between the probability of seeing no cars in a 20 minute window and the probability of seeing no cars in a 5 minute window. See that $$ P(N=0, t=20) = e^{-20 \lambda}, $$ $$ P(N=0, t=5) = e^{-5 \lambda}, $$ implies $$ P(N=0, t=20) = P(N=0, t=5)^4. $$ We know that $$ P(N=0, t=20) = 1 - P(N \geq 1, t=20) = \frac{16}{625} $$ and so $$ P(N=0, t=5) = \sqrt[4]{\frac{16}{625}} = \frac{2}{5} $$ and finally we come to our result. We calculate the probability of seeing a car in a 5 minute window as $$ P(N \geq 1, t=5) = 1 - P(N=0, t=5) = 1 - \frac{2}{5} = \frac{3}{5}. $$ The probability of seeing a car in a 5 minute window is $60\%$.