1. **Problem statement:** 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 $0.004$ of the time, so $P(\text{flagged} | L) = 0.004$ We want to find: (i) $P(S | \text{flagged})$ the probability an email is truly spam given it was flagged. (ii) $P(L | \text{not flagged})$ the probability an email is legitimate given it was not flagged. 2. **Formulas and rules:** We use Bayes' theorem: $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} $$ where $A$ and $B$ are events. Also, total probability rule: $$ P(B) = P(B|A)P(A) + P(B|A^c)P(A^c) $$ 3. **Step (i): Calculate $P(S | \text{flagged})$** - Calculate $P(\text{flagged})$: $$ P(\text{flagged}) = P(\text{flagged}|S)P(S) + P(\text{flagged}|L)P(L) = 1 \times 0.002 + 0.004 \times 0.998 $$ $$ = 0.002 + 0.003992 = 0.005992 $$ - Apply Bayes' theorem: $$ P(S|\text{flagged}) = \frac{P(\text{flagged}|S)P(S)}{P(\text{flagged})} = \frac{1 \times 0.002}{0.005992} $$ - Simplify: $$ P(S|\text{flagged}) = \frac{0.002}{0.005992} \approx 0.3336 $$ 4. **Step (ii): Calculate $P(L | \text{not flagged})$** - Calculate $P(\text{not flagged})$: $$ P(\text{not flagged}) = 1 - P(\text{flagged}) = 1 - 0.005992 = 0.994008 $$ - Calculate $P(\text{not flagged}|L)$: $$ P(\text{not flagged}|L) = 1 - P(\text{flagged}|L) = 1 - 0.004 = 0.996 $$ - Apply Bayes' theorem: $$ P(L|\text{not flagged}) = \frac{P(\text{not flagged}|L)P(L)}{P(\text{not flagged})} = \frac{0.996 \times 0.998}{0.994008} $$ - Simplify: $$ P(L|\text{not flagged}) = \frac{0.994008}{0.994008} = 1 $$ 5. **Interpretation:** - If an email is flagged, there is about a 33.36% chance it is truly spam. - If an email is not flagged, it is almost certainly legitimate (probability very close to 1). **Final answers:** (i) $P(S|\text{flagged}) \approx 0.3336$ (ii) $P(L|\text{not flagged}) \approx 1$