Miscellaneous · Q10: Ball moved between bags: the moved ball was black, $P = \frac{16}{31}$

Basics you need

Solution

1. Name the events

   Let

   • $E_1$⁠: the transferred ball is red;
   • $E_2$⁠: the transferred ball is black;
   • $A$⁠: the ball drawn from Bag II is red.

   $E_1$ and $E_2$ form a partition of the sample space: Bag I holds only red and black balls, so the moved ball is exactly one of the two colours.

   The colour moved is the hidden cause; the red draw is what we see. The question asks for the cause given what we see, so this is a Bayes' theorem problem.
   Partition of a sample space (Section 13.5.1). Lesson: Bayes' theorem

2. The first stage: which ball is moved

   Bag I has $3 + 4 = 7$ balls, so

   $$\begin{aligned} P(E_1) &= \frac37, \\ P(E_2) &= \frac47. \end{aligned}$$
   The ball is chosen at random from Bag I, so each of its 7 balls is equally likely to be moved.
   $P(E) = \frac{n(E)}{n(S)}$⁠. Basics: classical probability

3. The second stage: Bag II after the move

   Bag II had 4 red and 5 black balls, 9 in all; after the move it has 10.

   • If a red ball was moved ($E_1$⁠): 5 red, 5 black, so $P(A|E_1) = \frac{5}{10}$⁠.
   • If a black ball was moved ($E_2$⁠): 4 red, 6 black, so $P(A|E_2) = \frac{4}{10}$⁠.
   The moved ball joins Bag II before the draw, so it must be counted. Forgetting it would give $\frac49$ in both cases, and then the draw would tell us nothing about the moved ball.
   Conditional probability: the bag's contents are known in each case. Lesson: conditional probability

4. The chance of a red draw: total probability

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

   $$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac37\cdot\frac{5}{10} + \frac47\cdot\frac{4}{10} \\ &= \frac{15}{70} + \frac{16}{70} = \frac{31}{70}. \end{aligned}$$
   A red draw follows exactly one of the two moves, so the two path probabilities add.
   Multiply along a path, add the paths. Basics: tree diagrams

5. Black moved, given a red draw: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(E_2|A) &= \frac{P(E_2)P(A|E_2)}{P(A)} \\ &= \frac{16}{70} \div \frac{31}{70} \\ &= \frac{16}{70} \times \frac{70}{31} = \frac{16}{31}. \end{aligned}$$
   The top is the "black moved, then red drawn" path; the bottom is both paths to a red draw. 31 is prime and does not divide 16, so the fraction is in lowest terms.
   Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction

6. A quick sense check

   The other cause gets $P(E_1|A) = \frac{15}{70} \div \frac{31}{70} = \frac{15}{31}$⁠, and $\frac{16}{31} + \frac{15}{31} = 1$⁠. Before the draw, black was moved with probability $\frac47 \approx 0.571$⁠; the red draw lowers this to $\frac{16}{31} \approx 0.516$⁠.

   A red draw is less likely after a black ball is moved ($\frac{4}{10}$ against $\frac{5}{10}$⁠), so seeing red should make "black moved" less likely than before, but not by enough to fall below $\frac12$⁠.
   $P(E_1|A) + P(E_2|A) = 1$ for a partition. Lesson: Bayes' theorem

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