1. **State the problem:** We have a spam detection system with the following probabilities: - Probability an email is spam: $P(S) = 0.002$ - Probability an email is legitimate: $P(L) = 1 - P(S) = 0.998$ - The system always catches spam emails, so $P(\text{flagged}|S) = 1$ - The system falsely flags legitimate emails with probability $P(\text{flagged}|L) = 0.004$ We want to find: (i) $P(S|\text{flagged})$ the probability an email is truly spam given it is flagged. (ii) $P(L|\text{not flagged})$ the probability an email is legitimate given it is not flagged. 2. **Recall Bayes' theorem:** $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} $$ This allows us to find conditional probabilities. 3. **Calculate $P(\text{flagged})$:** $$ P(\text{flagged}) = P(\text{flagged}|S)P(S) + P(\text{flagged}|L)P(L) = 1 \times 0.002 + 0.004 \times 0.998 = 0.002 + 0.003992 = 0.005992 $$ 4. **Calculate $P(S|\text{flagged})$ using Bayes' theorem:** $$ P(S|\text{flagged}) = \frac{P(\text{flagged}|S)P(S)}{P(\text{flagged})} = \frac{1 \times 0.002}{0.005992} = \frac{0.002}{0.005992} $$ Simplify by dividing numerator and denominator: $$ P(S|\text{flagged}) = \frac{\cancel{0.002}}{\cancel{0.005992}} = 0.3338 \approx 0.334 $$ 5. **Calculate $P(\text{not flagged})$:** $$ P(\text{not flagged}) = 1 - P(\text{flagged}) = 1 - 0.005992 = 0.994008 $$ 6. **Calculate $P(L|\text{not flagged})$ using Bayes' theorem:** $$ P(L|\text{not flagged}) = \frac{P(\text{not flagged}|L)P(L)}{P(\text{not flagged})} $$ Where: $$ P(\text{not flagged}|L) = 1 - P(\text{flagged}|L) = 1 - 0.004 = 0.996 $$ So: $$ P(L|\text{not flagged}) = \frac{0.996 \times 0.998}{0.994008} = \frac{0.994008}{0.994008} = 1 $$ 7. **Interpretation:** - If an email is flagged, there is about a 33.4% chance it is truly spam. - If an email is not flagged, it is almost certainly legitimate (probability approximately 1). **Final answers:** (i) $P(S|\text{flagged}) \approx 0.334$ (ii) $P(L|\text{not flagged}) \approx 1$