1. ## independt variable probability

Professor stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particulare on the the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, also independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half of the crossings encountered.

A.) Which route should he take to minimize the probability of being late to the meeting?

B.) If he tosses a fair coin to decide on a route and is late, what is the probability he took the four crossing route?

2. I would use the binomial distribution formula

${{n}\choose{x}}*(p)^{x}*(1-P)^{n-x}$

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.

$pr(x\geq 1)=pr(x=1)+pr(x=2)$

[ ${{2}\choose{1}}*(0.1)^{1}*(1-0.1)^{2-1}$] + [ ${{2}\choose{2}}*(0.1)^{2}*(1-0.1)^{2-2}$]

$=0.18+0.01=0.19= 19\%$ 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.