Professor stan der Deviation can take one of two routes on his way home from wor

Question

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?

Explanation / Answer

The distributions are binomial with p=.1 and q=.9 Route with 4 crossing, with a probability of being stopped and the trains operate independently. probability of being late is(being stopped at 2 or more crossings P(x>=2)=1-P(x=0)-P(x=1)=1-C(4,0)(.1)^0(.9^4)-C(4,1)(.1)^1(.9)^3=1-.6561-.2916=.0523 ... Route with 2 crossings Probability of being stopped is (being stopped at 1 or more crossings) P(x>=1)=1-P(x=0)=1-c(2,0).1^0*(.9)^2=.19 Therefore chose the smaller probability of being late, the route with 4 crossings B) The probability he took the four crossing route is .5

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