I would use the binomial distribution formula

Where n= the total number of railroad crossings

p= the probability of being stopped at a crossing (0.1)

x=the number of railroad crossing that the professor gets stopped at.

So for route 1, if the professor is stopped at one more crossings he'll be late.

[ ] + [ ]

probability that the professor will be late if he takes route 1.

Now, just do the same thing for the second route.

I'll look at part b if I get a chance.