Exercise 13.3 · Q7: Insured drivers: a scooter driver, given an accident, $P = \frac{1}{52}$

Basics you need

Solution

1. Name the events

   Let

   • $E_1$⁠: the insured person is a scooter driver;
   • $E_2$⁠: the insured person is a car driver;
   • $E_3$⁠: the insured person is a truck driver;
   • $A$⁠: the insured person meets with an accident.

   $E_1$⁠, $E_2$⁠, $E_3$ form a partition of the sample space (all insured persons): the three groups do not overlap, together they make up all $12000$ insured people, and none is empty.

   The type of vehicle is the cause; the accident is the observed result. The question treats each insured person as belonging to exactly one of the three groups.
   Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem

2. The six probabilities as fractions

   There are $2000 + 4000 + 6000 = 12000$ insured people, so

   $$\begin{aligned} P(E_1) &= \frac{2000}{12000} = \frac16, \\ P(E_2) &= \frac{4000}{12000} = \frac13, \\ P(E_3) &= \frac{6000}{12000} = \frac12, \end{aligned}$$

   and the accident rates, converted from decimals, are

   $$\begin{aligned} P(A|E_1) &= 0.01 = \frac{1}{100}, \\ P(A|E_2) &= 0.03 = \frac{3}{100}, \\ P(A|E_3) &= 0.15 = \frac{15}{100}. \end{aligned}$$
   The person who has the accident is one of the insured, and before we know about the accident each insured person is equally likely, so a group's share of the 12000 is its probability. "The probability of an accidents are 0.01, 0.03 and 0.15 respectively" (the book's "an accidents" means "an accident") gives the accident chance inside each group, in the order scooter, car, truck. These are conditional probabilities.
   $P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: classical probability

3. The chance of an accident

   Since $E_1$⁠, $E_2$⁠, $E_3$ form a partition, by the theorem of total probability

   $$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac16\cdot\frac{1}{100} + \frac13\cdot\frac{3}{100} + \frac12\cdot\frac{15}{100} \\ &= \frac{1}{600} + \frac{1}{100} + \frac{15}{200} \\ &= \frac{1}{600} + \frac{6}{600} + \frac{45}{600} = \frac{52}{600}. \end{aligned}$$
   An accident happens to exactly one kind of driver, so the three path probabilities add. Keeping $\frac{52}{600}$ unsimplified makes the next step easy, because the 600 cancels.
   Multiply along a path, add the paths. Basics: tree diagrams

4. A scooter driver, given an accident: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{1}{600} \div \frac{52}{600} \\ &= \frac{1}{600} \times \frac{600}{52} = \frac{1}{52}. \end{aligned}$$
   The top is the scooter path; the bottom is all accident paths. Out of every 52 parts of accident probability, only 1 belongs to scooter drivers.
   Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction

5. A quick sense check

   For the other drivers, $P(E_2|A) = \frac{6}{52} = \frac{3}{26}$ and $P(E_3|A) = \frac{45}{52}$⁠. Then $\frac{1}{52} + \frac{6}{52} + \frac{45}{52} = 1$⁠.

   Scooter drivers are $\frac16$ of the insured but only $\frac{1}{52}$ of the accidents, because they have the lowest accident rate; truck drivers have $\frac{45}{52}$ of the accidents.

   Given an accident, it happened to exactly one kind of driver, so the three updated probabilities must add to 1.
   The posterior probabilities of a partition add to 1. Lesson: Bayes' theorem

Every fact used, and where it comes from