Example 20: Doctor arrives late: did he come by train? $P = \frac12$

Basics you need

Solution

1. Name the events

   Let $E$⁠: the doctor arrives late, and let $T_1$⁠, $T_2$⁠, $T_3$⁠, $T_4$⁠: he comes by train, bus, scooter and other means of transport, respectively.

   $T_1$⁠, $T_2$⁠, $T_3$⁠, $T_4$ form a partition of the sample space: on any visit he uses exactly one of them, and each has nonzero probability.

   The four ways of travelling are the causes; being late is the observed effect. The four probabilities $\frac{3}{10} + \frac15 + \frac{1}{10} + \frac25 = \frac{3 + 2 + 1 + 4}{10} = 1$ add to 1, which confirms that nothing is missing.
   A partition: pairwise disjoint, exhaustive, nonzero probabilities. Lesson: Bayes' theorem

2. The probabilities

   $$\begin{aligned} P(T_1) &= \frac{3}{10}, \quad P(T_2) = \frac15, \\ P(T_3) &= \frac{1}{10}, \quad P(T_4) = \frac25, \\ P(E|T_1) &= \frac14, \quad P(E|T_2) = \frac13, \\ P(E|T_3) &= \frac{1}{12}, \quad P(E|T_4) = 0. \end{aligned}$$
   The first four are given directly. "The probabilities that he will be late are $\frac14$⁠, $\frac13$⁠, and $\frac{1}{12}$⁠, if he comes by train, bus and scooter" are conditional on the way of travel, so they are $P(E|T_1)$⁠, $P(E|T_2)$⁠, $P(E|T_3)$⁠. "By other means ... he will not be late" means $P(E|T_4) = 0$⁠. (The book describes $P(E|T_1)$ in words as "Probability that the doctor arriving late comes by train", which is a description of $P(T_1|E)$⁠; the value $\frac14$ it uses is the chance of being late when he comes by train, $P(E|T_1)$⁠.)
   $P(E|T)$⁠: the chance of $E$ when $T$ is known to happen. Lesson: conditional probability

3. The chance of being late: total probability

   By the theorem of total probability,

   $$\begin{aligned} P(E) &= P(T_1)P(E|T_1) \\ &\quad + P(T_2)P(E|T_2) \\ &\quad + P(T_3)P(E|T_3) \\ &\quad + P(T_4)P(E|T_4) \\ &= \frac{3}{10}\cdot\frac14 + \frac15\cdot\frac13 \\ &\quad + \frac{1}{10}\cdot\frac{1}{12} + \frac25\cdot 0 \\ &= \frac{3}{40} + \frac{1}{15} + \frac{1}{120} + 0 \\ &= \frac{9}{120} + \frac{8}{120} + \frac{1}{120} \\ &= \frac{18}{120}. \end{aligned}$$
   He is late by exactly one way of travelling, so the four path probabilities add. With the common denominator 120: $\frac{3}{40} = \frac{9}{120}$⁠, $\frac{1}{15} = \frac{8}{120}$⁠. The term $\frac25 \cdot 0 = 0$ is written to show that the whole partition is used.
   Multiply along a path, add the paths. Basics: tree diagrams

4. By train, given late: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(T_1|E) &= \frac{P(T_1)P(E|T_1)}{P(E)} \\ &= \frac{9}{120} \div \frac{18}{120} = \frac{9}{18} = \frac12. \end{aligned}$$
   The top is the train path, $\frac{3}{10} \cdot \frac14 = \frac{3}{40} = \frac{9}{120}$⁠; the bottom is all late arrivals, $\frac{18}{120}$⁠. The 120 cancels. The book writes this as $\frac{3}{40} \times \frac{120}{18} = \frac12$⁠.
   Dividing by a fraction is multiplying by its flip. Basics: working with fractions

5. What the answer means

   Before he arrived, the train had probability $\frac{3}{10}$⁠. Knowing he is late raises it to $\frac12$⁠. The other ways get $P(T_2|E) = \frac{8}{18} = \frac49$⁠, $P(T_3|E) = \frac{1}{18}$ and $P(T_4|E) = 0$⁠; and $\frac{9 + 8 + 1 + 0}{18} = 1$⁠.

   "Other means" was the most likely way to travel ($\frac25$⁠), but it can never make him late, so once he is late it is ruled out completely. The train ends up the most likely way (9 of the 18 late arrivals) because it is both fairly likely and fairly often late. The bus is late most often ($\frac13$⁠), so its share grows the most, from $\frac15$ to $\frac49$⁠, more than double.
   For a partition, the $P(T_i|E)$ add to 1. Lesson: Bayes' theorem

Every fact used, and where it comes from