Example 22: Four boxes of coloured balls: black ball from box III, $P = \frac{156}{947}$

Basics you need

Solution

1. Name the events

   Let

   • $A$⁠: a black ball is drawn;
   • $E_1$⁠, $E_2$⁠, $E_3$⁠, $E_4$⁠: box I, II, III, IV is selected, respectively.

   $E_1, \ldots, E_4$ form a partition of the sample space: exactly one box is selected, and each can be.

   The box is the hidden cause, the black colour is the observed effect, and the question asks for the probability of the cause "box III" given the effect.
   A partition: pairwise disjoint, exhaustive, nonzero probabilities. Lesson: Bayes' theorem

2. The probabilities from the table

   Each box is selected with probability $P(E_i) = \frac14$⁠. Adding along each row, the boxes hold

   $$\begin{aligned} \text{I: } & 3 + 4 + 5 + 6 = 18, \\ \text{II: } & 2 + 2 + 2 + 2 = 8, \\ \text{III: } & 1 + 2 + 3 + 1 = 7, \\ \text{IV: } & 4 + 3 + 1 + 5 = 13 \end{aligned}$$

   balls, of which 3, 2, 1 and 4 are black. So

   $$\begin{aligned} P(A|E_1) &= \frac{3}{18}, \quad P(A|E_2) = \frac28, \\ P(A|E_3) &= \frac17, \quad P(A|E_4) = \frac{4}{13}. \end{aligned}$$
   "At random" gives the four boxes equal chances. Once a box is chosen, every ball in it is equally likely, so the chance of black is (black balls in the box) ÷ (all balls in the box). The boxes have different sizes, which is why the colour counts alone are not enough.
   $P = \frac{\text{favourable}}{\text{total}}$⁠. Basics: the classical definition

3. The chance of a black ball: total probability

   By the theorem of total probability,

   $$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &\quad + P(E_4)P(A|E_4) \\ &= \frac14\cdot\frac{3}{18} + \frac14\cdot\frac28 \\ &\quad + \frac14\cdot\frac17 + \frac14\cdot\frac{4}{13} \\ &= \frac{1}{24} + \frac{1}{16} + \frac{1}{28} + \frac{1}{13}. \end{aligned}$$

   The lowest common denominator is $4368 = 2^4 \times 3 \times 7 \times 13$⁠, so

   $$\begin{aligned} P(A) &= \frac{182}{4368} + \frac{273}{4368} \\ &\quad + \frac{156}{4368} + \frac{336}{4368} \\ &= \frac{947}{4368}. \end{aligned}$$
   The four black paths cannot happen together, so they add. Each product simplifies first: $\frac14\cdot\frac{3}{18} = \frac{3}{72} = \frac{1}{24}$⁠, $\frac14\cdot\frac28 = \frac{2}{32} = \frac{1}{16}$⁠, $\frac14\cdot\frac17 = \frac{1}{28}$⁠, $\frac14\cdot\frac{4}{13} = \frac{1}{13}$⁠. Then $4368 \div 24 = 182$⁠, $4368 \div 16 = 273$⁠, $4368 \div 28 = 156$ and $4368 \div 13 = 336$⁠.
   Multiply along a path, add the paths. Basics: tree diagrams

4. Box III, given black: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(E_3|A) &= \frac{P(E_3)P(A|E_3)}{P(A)} \\ &= \frac{156}{4368} \div \frac{947}{4368} \\ &= \frac{156}{947} \approx 0.165. \end{aligned}$$
   The top is the box III path, $\frac{1}{28} = \frac{156}{4368}$⁠; the bottom is all black draws. The common denominator cancels. 947 is prime, so $\frac{156}{947}$ does not simplify. The book writes the same fraction and gives its value as 0.165; the exact value is $\frac{156}{947}$⁠, and $0.16473\ldots$ rounds to 0.165.
   Bayes' theorem: one path over all the paths. Lesson: Bayes' theorem

5. A shortcut: the $\frac14$ cancels

   Every box has the same $P(E_i) = \frac14$⁠, so it cancels from the top and the bottom of Bayes' formula:

   $$\begin{aligned} P(E_3|A) &= \frac{P(A|E_3)}{\sum_{i=1}^{4} P(A|E_i)} \\ &= \frac17 \div \Big(\frac{3}{18} + \frac28 + \frac17 + \frac{4}{13}\Big) \\ &= \frac{156}{1092} \div \frac{947}{1092} = \frac{156}{947}. \end{aligned}$$
   With equal priors, each box's posterior is just its own $P(A|E_i)$ divided by the sum of all of them. Here $\frac{3}{18} + \frac28 + \frac17 + \frac{4}{13} = \frac{182 + 273 + 156 + 336}{1092} = \frac{947}{1092}$⁠.
   A factor common to every term cancels. Basics: working with fractions

Every fact used, and where it comes from