1. **Problem statement:** You spin a wheel with two letters: A (3/4 probability) and B (1/4 probability). Then you draw a shape from the corresponding drum (Lostrommel). Given that the second draw is a triangle, find the probability that the wheel showed "B". 2. **Known data:** - Glücksrad probabilities: $P(A) = \frac{3}{4}$, $P(B) = \frac{1}{4}$. - Lostrommel A contains triangles numbered 1,2,3,4 and circles 5-9. - Lostrommel B contains squares 1,4,5,6 and circles 2,3,7,8,9. 3. **Calculate probability of drawing a triangle given A or B:** - Lostrommel A: triangles = 4 shapes, total shapes = 9, so $P(\text{triangle}|A) = \frac{4}{9}$. - Lostrommel B: triangles = 0 (no triangles), so $P(\text{triangle}|B) = 0$. 4. **Use Bayes' theorem:** $$ P(B|\text{triangle}) = \frac{P(\text{triangle}|B) \cdot P(B)}{P(\text{triangle}|A) \cdot P(A) + P(\text{triangle}|B) \cdot P(B)} $$ 5. **Substitute values:** $$ P(B|\text{triangle}) = \frac{0 \cdot \frac{1}{4}}{\frac{4}{9} \cdot \frac{3}{4} + 0 \cdot \frac{1}{4}} = 0 $$ 6. **Interpretation:** Since Lostrommel B has no triangles, if a triangle is drawn, the wheel could not have shown B. --- **Final answer:** $$ P(B|\text{triangle}) = 0 $$