1. **Problem statement:** A group of office workers were surveyed. Given: - Probability of taking regular exercise $P(R) = \frac{2}{5}$. - Probability of always eating breakfast $P(E) = \frac{2}{3}$. - Probability of taking regular exercise given always eating breakfast $P(R|E) = \frac{9}{25}$. Find: (a) $P(E \cap R)$ (b) $P(E^c \cap R^c)$ (c) Whether $E$ and $R$ are statistically independent. 2. **Formulas and rules:** - Joint probability: $P(E \cap R) = P(E) \times P(R|E)$. - Complement rule: $P(E^c) = 1 - P(E)$ and $P(R^c) = 1 - P(R)$. - For independence: $P(E \cap R) = P(E) \times P(R)$ must hold. 3. **Calculate (a) $P(E \cap R)$:** $$ P(E \cap R) = P(E) \times P(R|E) = \frac{2}{3} \times \frac{9}{25} = \frac{18}{75} = \frac{6}{25} $$ 4. **Calculate (b) $P(E^c \cap R^c)$:** First find complements: $$ P(E^c) = 1 - \frac{2}{3} = \frac{1}{3}, \quad P(R^c) = 1 - \frac{2}{5} = \frac{3}{5} $$ We need $P(E^c \cap R^c)$. Use the formula: $$ P(E^c \cap R^c) = 1 - P(E \cup R) = 1 - [P(E) + P(R) - P(E \cap R)] $$ Calculate inside the bracket: $$ P(E) + P(R) - P(E \cap R) = \frac{2}{3} + \frac{2}{5} - \frac{6}{25} = \frac{50}{75} + \frac{30}{75} - \frac{18}{75} = \frac{62}{75} $$ So: $$ P(E^c \cap R^c) = 1 - \frac{62}{75} = \frac{13}{75} $$ 5. **Check (c) independence:** Calculate $P(E) \times P(R)$: $$ \frac{2}{3} \times \frac{2}{5} = \frac{4}{15} = \frac{20}{75} $$ Compare with $P(E \cap R) = \frac{6}{25} = \frac{18}{75}$. Since $\frac{18}{75} \neq \frac{20}{75}$, $E$ and $R$ are **not independent**. **Final answers:** (a) $P(E \cap R) = \frac{6}{25}$ (b) $P(E^c \cap R^c) = \frac{13}{75}$ (c) $E$ and $R$ are not statistically independent because $P(E \cap R) \neq P(E) \times P(R)$.