Let's solve Question 1 step by step.

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Part a:

This is a binomial distribution problem, where:

• $n = 11$ (number of trials, i.e., stolen cars reported),
• $p = 0.35$ (probability of success, i.e., a car is found),
• $q = 1 - p = 0.65$ (probability of failure).

The binomial probability formula is: $P(X = k) = \binom{n}{k} p^k q^{n-k},$ where $\binom{n}{k}$ is the binomial coefficient.

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(i) Exactly three stolen cars will be found in a month:

$P(X = 3) = \binom{11}{3} (0.35)^3 (0.65)^8$ First, calculate the binomial coefficient: $\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165.$ Now substitute into the formula: $P(X = 3) = 165 \times (0.35)^3 \times (0.65)^8.$ Numerical result: $P(X = 3) \approx 165 \times 0.042875 \times 0.041652 \approx 0.293.$

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(ii) Not more than two stolen cars will be found in a month:

This is the cumulative probability: $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2).$

• For $X = 0$: $P(X = 0) = \binom{11}{0} (0.35)^0 (0.65)^{11} = (0.65)^{11} \approx 0.0137.$

• For $X = 1$: $P(X = 1) = \binom{11}{1} (0.35)^1 (0.65)^{10} = 11 \times 0.35 \times (0.65)^{10} \approx 0.0809.$

• For $X = 2$: $P(X = 2) = \binom{11}{2} (0.35)^2 (0.65)^9 = \frac{11 \times 10}{2} \times (0.35)^2 \times (0.65)^9 \approx 55 \times 0.1225 \times 0.0271 \approx 0.183.$

Summing up: $P(X \leq 2) = 0.0137 + 0.0809 + 0.183 \approx 0.2776.$

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Part b:

This is a Poisson distribution problem, where:

• The average rate $\lambda = 3$ (books per year).

The Poisson probability formula is: $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}.$

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(i) Exactly three books in a given year:

$P(X = 3) = \frac{3^3 e^{-3}}{3!} = \frac{27 e^{-3}}{6}.$ Using $e^{-3} \approx 0.0498$: $P(X = 3) \approx \frac{27 \times 0.0498}{6} \approx 0.224.$

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(ii) At least one book in a two-year period:

For two years, the average rate $\lambda = 6$. The complement of "at least one book" is "no books": $P(X \geq 1) = 1 - P(X = 0),$ where: $P(X = 0) = \frac{6^0 e^{-6}}{0!} = e^{-6}.$ Using $e^{-6} \approx 0.00248$: $P(X \geq 1) = 1 - 0.00248 \approx 0.9975.$

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(iii) Less than three books in half a year:

For half a year, the average rate $\lambda = 1.5$. $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).$

• For $X = 0$: $P(X = 0) = \frac{1.5^0 e^{-1.5}}{0!} = e^{-1.5} \approx 0.2231.$

• For $X = 1$: $P(X = 1) = \frac{1.5^1 e^{-1.5}}{1!} = 1.5 \times e^{-1.5} \approx 0.3346.$

• For $X = 2$: [ P(X = 2) = \frac{1.5^2 e^{-1.5}}{2!} = \frac{2.25 \times e^{-1.5}}{2