Question

A commuter can take any one of two routes on his way home from work. On the ¯rst route,
there are four railroad crossings. The probability that he will be stopped by a train at any
particular one of the crossings is .1, and trains operate independently at the four crossings.
The other route is longer, but there are only two crossing, independent of one another, with
the same stoppage probability for each as on the ¯rst route. On a particular day, the
commuter 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 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 he is late, what is the probability that
he took the four-crossing route?

Answer 1

Let's calculate probability of being late at first and second routes.

Probability of being late at first route:

$$P _ {1} = 1 - 0. 9 ^ {4} - 4 \cdot 0. 1 \cdot 0. 9 ^ {3} = 0. 0 5 2 3$$

Probability of being late at second route:

$$P _ {2} = 1 - 0. 9 ^ {2} = 0. 1 9$$

So, it is better to choose the first route.

b)

$$\begin{array}{l} P (\text {first route} | \text {late}) = (\text {using Bayers formula}) = \frac {P (\text {late} | \text {first route}) P (\text {first route})}{P (\text {late})} \\ = \frac {0 . 0 5 2 3 \cdot \frac {1}{2}}{\frac {1}{2} \cdot 0 . 0 5 2 3 + \frac {1}{2} \cdot 0 . 1 9} = \frac {5 2 3}{2 4 2 3} \\ \end{array}$$