Question:
<p>Box I: 3 white, 2 black; Box II: 2 white, 3 black; Box III: 2 white, 2 black. A box is selected at random and one ball is drawn at random. Given the ball is white, which are TRUE?</p>
Step-by-Step Solution
Key Concept: Apply Bayes' theorem. P(white) = (1/3)(3/5 + 2/5 + 2/4). Compare posterior probabilities.
<p>P(W|I)=3/5, P(W|II)=2/5, P(W|III)=2/4=1/2. P(box)=1/3 each.</p><p>$P(W) = \frac{1}{3}\left(\frac{3}{5}+\frac{2}{5}+\frac{1}{2}\right) = \frac{1}{3}\cdot\frac{6+4+5}{10} = \frac{15}{30} = \frac{1}{2}$.</p><p>Posteriors (proportional to likelihoods): I: 3/5, II: 2/5, III: 1/2.</p><p>P(I|W) = (3/5)/(3/2) = 2/5; P(II|W) = (2/5)/(3/2)=4/15; P(III|W)=(1/2)/(3/2)=1/3.</p><p>A: 2/5 > 4/15 ✓ B: 1/3 > 4/15 ✓ C: 2/5 ≠ 6/17 ✗ D: 4/15 ≠ 1/3 ✗</p><p>Answer: AB ✓</p>
Correct Answer: AB