Exercise 13.3 · Q13 (MCQ): Speaks the truth with $\frac45$⁠: head actually appeared, $P = \frac45$

Basics you need

Solution

1. Name the events

   Let

   • $E_1$⁠: a head appears on the coin;
   • $E_2$⁠: a tail appears on the coin;
   • $R$⁠: A reports that a head appears.

   $E_1$ and $E_2$ form a partition of the sample space: a toss gives exactly one of head and tail, and each is possible.

   What the coin really shows is the hidden cause; A's report is the result we observe. We write $R$ for the report because the letter A already names the person.
   Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem

2. The four probabilities

   $$\begin{aligned} P(E_1) &= \frac12, \\ P(E_2) &= \frac12, \end{aligned}$$ $$\begin{aligned} P(R|E_1) &= \frac45, \\ P(R|E_2) &= 1 - \frac45 = \frac15. \end{aligned}$$
   The coin is taken as fair, as usual when nothing else is said. If a head appeared, A reports a head exactly when A tells the truth. If a tail appeared, A reports a head exactly when A lies. A's chance of telling the truth is taken to be $\frac45$ whatever the coin shows.
   $P(\text{lie}) = 1 - P(\text{truth})$⁠. Basics: rules of probability

3. The chance that A reports a head

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

   $$\begin{aligned} P(R) &= P(E_1)P(R|E_1) \\ &\quad + P(E_2)P(R|E_2) \\ &= \frac12\cdot\frac45 + \frac12\cdot\frac15 \\ &= \frac{4}{10} + \frac{1}{10} = \frac{5}{10}. \end{aligned}$$
   A report of a head comes either from a head and the truth or from a tail and a lie, never both, so the two path probabilities add.
   Multiply along a path, add the paths. Basics: tree diagrams

4. A head, given the report: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(E_1|R) &= \frac{P(E_1)P(R|E_1)}{P(R)} \\ &= \frac{4}{10} \div \frac{5}{10} \\ &= \frac{4}{10} \times \frac{10}{5} = \frac45. \end{aligned}$$
   The top is the "true head" path; the bottom is both routes to the report. The answer equals A's chance of telling the truth only because the coin is fair: with $P(E_1) = P(E_2)$⁠, the halves cancel and $\frac{\frac45}{\frac45 + \frac15} = \frac45$⁠.
   Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction

5. Examine every option

   • (A) $\frac45$⁠: equals $P(E_1|R)$ from Step 4. Correct.
   • (B) $\frac12$⁠: this is $P(E_1)$⁠, the chance of a head before the report is heard. It ignores the report, which is evidence for a head. Wrong.
   • (C) $\frac15$⁠: this is $P(R|E_2)$⁠, the chance that A lies; it is also $P(E_2|R) = 1 - \frac45$⁠, the chance that there was a tail after all. Wrong.
   • (D) $\frac25$⁠: this is $\frac12\cdot\frac45 = \frac{4}{10}$⁠, the probability of the path "head and a true report", $P(E_1 \cap R)$⁠. It forgets to divide by $P(R) = \frac{5}{10}$⁠. Wrong.
   Each wrong option is a number that appears on the way to the answer, which is why it is offered. Only the conditional probability $P(E_1|R)$ answers "given that A reports a head".
   $P(E_1|R) = \frac{P(E_1 \cap R)}{P(R)}$⁠. Lesson: conditional probability

Every fact used, and where it comes from