1. **Problem statement:** We are given the probability that a car is stolen overnight in an unsafe area is $p=0.1$. There are $n=12$ cars parked. We want to find probabilities for different numbers of stolen cars. 2. **Formula:** This is a binomial probability problem. The probability of exactly $k$ successes (cars stolen) in $n$ trials is given by: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 3. **(i) No car is stolen:** This means $k=0$. $$P(X=0) = \binom{12}{0} (0.1)^0 (0.9)^{12} = 1 \times 1 \times 0.9^{12} = 0.9^{12}$$ Calculate $0.9^{12} \approx 0.2824$. 4. **(ii) At most two cars are stolen:** This means $k=0,1,2$. $$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$$ Calculate each term: $$P(X=1) = \binom{12}{1} (0.1)^1 (0.9)^{11} = 12 \times 0.1 \times 0.9^{11}$$ $$P(X=2) = \binom{12}{2} (0.1)^2 (0.9)^{10} = 66 \times 0.01 \times 0.9^{10}$$ Calculate values: $0.9^{11} \approx 0.3138$, $0.9^{10} \approx 0.3487$ So, $$P(X=1) = 12 \times 0.1 \times 0.3138 = 0.3766$$ $$P(X=2) = 66 \times 0.01 \times 0.3487 = 0.2302$$ Sum: $$P(X \leq 2) = 0.2824 + 0.3766 + 0.2302 = 0.8892$$ 5. **(iii) At least nine cars are stolen:** This means $k \geq 9$. Calculate: $$P(X \geq 9) = P(X=9) + P(X=10) + P(X=11) + P(X=12)$$ Calculate each term: $$P(X=9) = \binom{12}{9} (0.1)^9 (0.9)^3 = 220 \times 10^{-9} \times 0.729 = 1.6038 \times 10^{-7}$$ $$P(X=10) = \binom{12}{10} (0.1)^{10} (0.9)^2 = 66 \times 10^{-10} \times 0.81 = 5.346 \times 10^{-8}$$ $$P(X=11) = \binom{12}{11} (0.1)^{11} (0.9)^1 = 12 \times 10^{-11} \times 0.9 = 1.08 \times 10^{-8}$$ $$P(X=12) = \binom{12}{12} (0.1)^{12} (0.9)^0 = 1 \times 10^{-12} \times 1 = 10^{-12}$$ Sum: $$P(X \geq 9) \approx 1.6038 \times 10^{-7} + 5.346 \times 10^{-8} + 1.08 \times 10^{-8} + 10^{-12} \approx 2.246 \times 10^{-7}$$ **Final answers:** - (i) $P(X=0) \approx 0.2824$ - (ii) $P(X \leq 2) \approx 0.8892$ - (iii) $P(X \geq 9) \approx 0.0000002246$