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?

A) He should take the route with 4 crossings

B) The probability that he took the 4 crossing route is 0.2158

Step-by-step explanation:

Lets call X the number of crossings he encounters, A if he takes route 1 and B if he takes route 2.

Note that X given A is a binomial random variable with parameters n = 4 p = 0.1, and X given B has parameters n = 2, p = 0.1

The probability that the Professor is on time on route 1 is equal to

P(X|A = 0) + P(X|A = 1) = 0.9⁴ + 0.9³*0.1*4 = 0.9477

On the other hand, the probability that the professor is on time on route 2 is

P(X|B = 0) = 0.9² = 0.81

Hence, it is more likely for the professor to be late on route 2, thus he should take the route 1, the one with 4 crossings.

B) Lets call L the event 'The professor is late'. We know that

P(L|A) = 1-0.9477 = 0.0524

P(L|B) = 1-0.81 = 0.19

Also

P(A) = P(B) = 1/2 (this only depends on the result of the coin.

For the Bayes theorem we know, therefore that

Hence, the probability that he took the 4 crossing route is 0.2158.

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4