1. **Problem 1: Probability of raining on exactly one of the next two days** We are given: - If it rains on a day, probability it rains next day = $\frac{7}{10}$ - If it does not rain on a day, probability it rains next day = $\frac{4}{10}$ - It is raining today. We want the probability it rains on exactly one of the next two days. 2. **Tree diagram setup:** - Day 1 (tomorrow): Rains (R) with probability $\frac{7}{10}$, No rain (N) with probability $\frac{3}{10}$ - Day 2 (day after tomorrow): - If Day 1 = R, then Day 2 = R with $\frac{7}{10}$, N with $\frac{3}{10}$ - If Day 1 = N, then Day 2 = R with $\frac{4}{10}$, N with $\frac{6}{10}$ 3. **Calculate probabilities for exactly one rainy day in next two days:** - Case 1: Rain on Day 1, No rain on Day 2 $$P(R,N) = \frac{7}{10} \times \frac{3}{10} = \frac{21}{100}$$ - Case 2: No rain on Day 1, Rain on Day 2 $$P(N,R) = \frac{3}{10} \times \frac{4}{10} = \frac{12}{100}$$ 4. **Add these probabilities:** $$P(\text{exactly one rain}) = \frac{21}{100} + \frac{12}{100} = \frac{33}{100} = 0.33$$ --- 5. **Problem 2: Probability outcome is a prime number on biased dice** Given probabilities: - 1: 0.2 - 2: 0.17 - 3: 0.18 - 4: 0.15 - 5: unknown - 6: 0.03 Sum of known probabilities: $$0.2 + 0.17 + 0.18 + 0.15 + 0.03 = 0.73$$ Probability for 5: $$1 - 0.73 = 0.27$$ Prime numbers on dice: 2, 3, 5 Sum their probabilities: $$0.17 + 0.18 + 0.27 = 0.62$$ --- 6. **Problem 3: Distance from starting position after two legs of a journey** Boat travels: - 11.2 km at bearing 055° - 9.8 km at bearing 275° We want the straight-line distance from start. 7. **Convert bearings to vectors:** - Bearing 055° means angle 55° from north clockwise. - Bearing 275° means angle 275° from north clockwise. 8. **Calculate components (using north as y-axis, east as x-axis):** - First leg: $$x_1 = 11.2 \times \sin(55^\circ)$$ $$y_1 = 11.2 \times \cos(55^\circ)$$ - Second leg: $$x_2 = 9.8 \times \sin(275^\circ)$$ $$y_2 = 9.8 \times \cos(275^\circ)$$ 9. **Calculate total displacement components:** $$x = x_1 + x_2$$ $$y = y_1 + y_2$$ 10. **Distance from start:** $$d = \sqrt{x^2 + y^2}$$ 11. **Numerical values:** - $\sin(55^\circ) \approx 0.8192$, $\cos(55^\circ) \approx 0.5736$ - $\sin(275^\circ) = \sin(275^\circ - 360^\circ) = \sin(-85^\circ) \approx -0.9962$ - $\cos(275^\circ) = \cos(-85^\circ) \approx 0.0872$ Calculate: $$x = 11.2 \times 0.8192 + 9.8 \times (-0.9962) = 9.172 + (-9.761) = -0.589$$ $$y = 11.2 \times 0.5736 + 9.8 \times 0.0872 = 6.423 + 0.854 = 7.277$$ Distance: $$d = \sqrt{(-0.589)^2 + 7.277^2} = \sqrt{0.347 + 52.95} = \sqrt{53.297} \approx 7.3$$ **Final answer:** The boat is approximately 7.3 km from its starting position.