1. **State the problem:** We have a circular archery target divided into four parts by concentric circles with radii 3 cm, 9 cm, 15 cm, and 30 cm. The scoring regions are: - Central circle (radius 3 cm) scores 5 points - Ring between 3 cm and 9 cm scores 3 points - Ring between 9 cm and 15 cm scores 2 points - Ring between 15 cm and 30 cm scores 1 point We assume an arrow hits exactly one region, equally likely anywhere on the target. 2. **Calculate areas of each region:** - Area of circle with radius $r$ is $\pi r^2$. - Area of central circle (5 points): $$A_5 = \pi \times 3^2 = 9\pi$$ - Area of ring for 3 points (between 3 and 9 cm): $$A_3 = \pi \times 9^2 - \pi \times 3^2 = 81\pi - 9\pi = 72\pi$$ - Area of ring for 2 points (between 9 and 15 cm): $$A_2 = \pi \times 15^2 - \pi \times 9^2 = 225\pi - 81\pi = 144\pi$$ - Area of ring for 1 point (between 15 and 30 cm): $$A_1 = \pi \times 30^2 - \pi \times 15^2 = 900\pi - 225\pi = 675\pi$$ 3. **Calculate total area:** $$A_{total} = \pi \times 30^2 = 900\pi$$ 4. **a) Probability of scoring 5 points:** $$P(5) = \frac{A_5}{A_{total}} = \frac{9\pi}{900\pi} = \frac{9}{900} = 0.01$$ 5. **b) Probabilities of scoring 3, 2, and 1 points:** - $$P(3) = \frac{A_3}{A_{total}} = \frac{72\pi}{900\pi} = \frac{72}{900} = 0.08$$ - $$P(2) = \frac{A_2}{A_{total}} = \frac{144\pi}{900\pi} = \frac{144}{900} = 0.16$$ - $$P(1) = \frac{A_1}{A_{total}} = \frac{675\pi}{900\pi} = \frac{675}{900} = 0.75$$ 6. **c) Probability of scoring 1 point given no 5 points:** - Probability of no 5 points: $$P(\text{not }5) = 1 - P(5) = 1 - 0.01 = 0.99$$ - Probability of scoring 1 point given no 5 points: $$P(1|\text{not }5) = \frac{P(1)}{P(\text{not }5)} = \frac{0.75}{0.99} \approx 0.7576$$ 7. **d) Probability that neither of two arrows scores 1 point given total score 6:** - Possible scoring pairs for total 6 points: - (5,1), (1,5), (3,3) - Probability of each pair: - $$P(5)P(1) = 0.01 \times 0.75 = 0.0075$$ - $$P(1)P(5) = 0.75 \times 0.01 = 0.0075$$ - $$P(3)P(3) = 0.08 \times 0.08 = 0.0064$$ - Total probability of scoring 6 points: $$P(6) = 0.0075 + 0.0075 + 0.0064 = 0.0214$$ - Probability that neither arrow scores 1 point means both arrows score 3 points: $$P(\text{no }1|6) = \frac{P(3)P(3)}{P(6)} = \frac{0.0064}{0.0214} \approx 0.2991$$ **Final answers:** - a) $P(5) = 0.01$ - b) $P(3) = 0.08$, $P(2) = 0.16$, $P(1) = 0.75$ - c) $P(1|\text{not }5) \approx 0.7576$ - d) $P(\text{no }1|6) \approx 0.2991$