Problem Statement
Bag I contains $3$ red and $4$ black balls and Bag II contains $5$ red and $6$ black balls. One ball is drawn at random from Bag I and transferred to Bag II. A ball is then drawn from Bag II and found to be red. What is the probability that the transferred ball was black?
Verified Solution & Marking Scheme
Define Transfer Events and Priors
Let $E_1$: Transferred ball from Bag I was Red $\implies P(E_1) = \frac{3}{7}$.
Let $E_2$: Transferred ball from Bag I was Black $\implies P(E_2) = \frac{4}{7}$.
Let $A$: Ball drawn from Bag II is Red.
Determine Conditional Probabilities for Drawing Red from Bag II
Bag II originally has 5 red, 6 black (11 total).
- If red transferred ($E_1$): Bag II has 6 red, 6 black (12 total) $\implies P(A|E_1) = \frac{6}{12} = \frac{1}{2}$.
- If black transferred ($E_2$): Bag II has 5 red, 7 black (12 total) $\implies P(A|E_2) = \frac{5}{12}$.
Apply Bayes' Theorem for P(E₂|A)
$P(E_2|A) = \frac{P(E_2) P(A|E_2)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2)}$
$= \frac{\frac{4}{7} \times \frac{5}{12}}{\frac{3}{7} \times \frac{6}{12} + \frac{4}{7} \times \frac{5}{12}} = \frac{\frac{20}{84}}{\frac{18 + 20}{84}} = \frac{20}{38} = \frac{10}{19}$