1. **Problem statement:** The probability it will rain on any day next week is half the probability it will rain on the same day this week; otherwise, it remains the same. Given: Probability it rains on Friday this week $= \frac{1}{4}$. Find: (i) Probability it will not rain next Friday. (ii) Probability it will rain on exactly two Fridays out of three consecutive Fridays starting this week. 2. **Step (i): Probability it will not rain next Friday** - Probability it rains next Friday $= \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$. - Probability it does not rain next Friday $= 1 - \frac{1}{8} = \frac{7}{8}$. 3. **Step (ii): Probability it rains on exactly two Fridays out of three consecutive Fridays starting this week** - Let $p$ be the probability it rains on Friday this week $= \frac{1}{4}$. - Probability it rains next Friday $= \frac{1}{8}$. - Probability it rains the Friday after next $= \frac{1}{8}$ (same as next Friday). 4. **Calculate the probability of exactly two rainy Fridays out of these three:** - Possible cases: 1. Rain this week and next week, no rain Friday after next: $p \times \frac{1}{2}p \times (1 - \frac{1}{2}p)$ 2. Rain this week and Friday after next, no rain next week: $p \times (1 - \frac{1}{2}p) \times \frac{1}{2}p$ 3. Rain next week and Friday after next, no rain this week: $(1 - p) \times \frac{1}{2}p \times \frac{1}{2}p$ - Substitute $p = \frac{1}{4}$: 1. $\frac{1}{4} \times \frac{1}{8} \times (1 - \frac{1}{8}) = \frac{1}{4} \times \frac{1}{8} \times \frac{7}{8} = \frac{7}{256}$ 2. $\frac{1}{4} \times (1 - \frac{1}{8}) \times \frac{1}{8} = \frac{1}{4} \times \frac{7}{8} \times \frac{1}{8} = \frac{7}{256}$ 3. $(1 - \frac{1}{4}) \times \frac{1}{8} \times \frac{1}{8} = \frac{3}{4} \times \frac{1}{8} \times \frac{1}{8} = \frac{3}{256}$ - Total probability $= \frac{7}{256} + \frac{7}{256} + \frac{3}{256} = \frac{17}{256}$. 5. **Check given answer:** Given answer is $\frac{25}{312}$ which simplifies approximately to $0.0801$. Our calculation $\frac{17}{256} \approx 0.0664$. 6. **Recalculate using binomial probability:** - Probabilities for each Friday: $p_1 = \frac{1}{4}$, $p_2 = \frac{1}{8}$, $p_3 = \frac{1}{8}$. - Probability of rain on exactly two Fridays: $$P = p_1 p_2 (1-p_3) + p_1 (1-p_2) p_3 + (1-p_1) p_2 p_3$$ - Substitute values: $$= \frac{1}{4} \times \frac{1}{8} \times \left(1 - \frac{1}{8}\right) + \frac{1}{4} \times \left(1 - \frac{1}{8}\right) \times \frac{1}{8} + \left(1 - \frac{1}{4}\right) \times \frac{1}{8} \times \frac{1}{8}$$ $$= \frac{1}{4} \times \frac{1}{8} \times \frac{7}{8} + \frac{1}{4} \times \frac{7}{8} \times \frac{1}{8} + \frac{3}{4} \times \frac{1}{8} \times \frac{1}{8}$$ $$= \frac{7}{256} + \frac{7}{256} + \frac{3}{256} = \frac{17}{256}$$ 7. **Final answers:** (i) Probability it will not rain next Friday $= \frac{7}{8}$. (ii) Probability it will rain on exactly two Fridays out of three $= \frac{17}{256}$. Note: The given answer $\frac{25}{312}$ does not match the calculation based on the problem statement.