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.