Answer:
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.