Question: CONSOLIDATION AND EXTENSION EXERCISE (a) In a certain population 10% of the people smoke. It has been observed that 2% of the people in the population develop lung cancer. 1% of people in the population smoke and develop lung cancer. Are ‘smoking’ and ‘developing lung cancer’ independent? Motivate. (b) A and B are such that P(not A) = 0,48 and P(B) = 0,25. Determine P(A or B) if (1) A and B are mutually exclusive. (2) A and B are independent. (c) Given: P(M) = 0,45, P(N) = 0,3 and P(M or N) = 0,615. (1) Determine (i) P(M and N) (ii) P(M and (not N)) (iii) P((not M) or (not N)) (2) Are the events M and N independent? Motivate. (d) In a city 10% of all crimes are violent. 90% of violent crimes are reported, but only 50% of non-violent crimes are reported. What is the probability that a random crime will be reported? (e) In a small village with 10 000 inhabitants, people are divided into 2 tribes:

1. **Problem Statement:** (a) Given: - $P(\text{smoke}) = 0.10$ - $P(\text{lung cancer}) = 0.02$ - $P(\text{smoke and lung cancer}) = 0.01$ Determine if smoking and developing lung cancer are independent. 2. **Formula for Independence:** Two events $A$ and $B$ are independent if and only if: $$P(A \cap B) = P(A) \times P(B)$$ 3. **Check Independence:** Calculate $P(\text{smoke}) \times P(\text{lung cancer})$: $$0.10 \times 0.02 = 0.002$$ 4. **Compare with Given:** Given $P(\text{smoke and lung cancer}) = 0.01$ which is not equal to $0.002$. 5. **Conclusion:** Since $P(\text{smoke and lung cancer}) \neq P(\text{smoke}) \times P(\text{lung cancer})$, smoking and developing lung cancer are **not independent**. --- 1. **Problem Statement:** Given: - $P(\neg A) = 0.48 \Rightarrow P(A) = 1 - 0.48 = 0.52$ - $P(B) = 0.25$ Find $P(A \cup B)$ for: (1) $A$ and $B$ mutually exclusive (2) $A$ and $B$ independent 2. **Formula for Union:** $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ 3. **Case (1) Mutually Exclusive:** $$P(A \cap B) = 0$$ So, $$P(A \cup B) = 0.52 + 0.25 - 0 = 0.77$$ 4. **Case (2) Independent:** $$P(A \cap B) = P(A) \times P(B) = 0.52 \times 0.25 = 0.13$$ So, $$P(A \cup B) = 0.52 + 0.25 - 0.13 = 0.64$$ --- 1. **Problem Statement:** Given: - $P(M) = 0.45$ - $P(N) = 0.3$ - $P(M \cup N) = 0.615$ 2. **Formula for Union:** $$P(M \cup N) = P(M) + P(N) - P(M \cap N)$$ 3. **Calculate $P(M \cap N)$:** $$P(M \cap N) = P(M) + P(N) - P(M \cup N) = 0.45 + 0.3 - 0.615 = 0.135$$ 4. **Calculate $P(M \cap \neg N)$:** $$P(M \cap \neg N) = P(M) - P(M \cap N) = 0.45 - 0.135 = 0.315$$ 5. **Calculate $P(\neg M \cup \neg N)$:** Using De Morgan's law: $$P(\neg M \cup \neg N) = 1 - P(M \cap N) = 1 - 0.135 = 0.865$$ 6. **Check Independence:** For independence: $$P(M \cap N) = P(M) \times P(N) = 0.45 \times 0.3 = 0.135$$ 7. **Conclusion:** Since $P(M \cap N) = P(M) \times P(N)$, events $M$ and $N$ are **independent**. --- 1. **Problem Statement:** Given: - $P(\text{violent crime}) = 0.10$ - $P(\text{reported} | \text{violent}) = 0.90$ - $P(\text{reported} | \text{non-violent}) = 0.50$ 2. **Calculate $P(\text{non-violent crime})$:** $$P(\text{non-violent}) = 1 - 0.10 = 0.90$$ 3. **Use Law of Total Probability:** $$P(\text{reported}) = P(\text{reported} | \text{violent}) \times P(\text{violent}) + P(\text{reported} | \text{non-violent}) \times P(\text{non-violent})$$ 4. **Calculate:** $$P(\text{reported}) = 0.90 \times 0.10 + 0.50 \times 0.90 = 0.09 + 0.45 = 0.54$$ 5. **Conclusion:** The probability that a random crime will be reported is $0.54$. --- (e) Problem statement incomplete; no calculation performed.