To maintain the good condition of library books, any book with more than 7 defects will be donated away. The mean number of defects per book in the reference section is 3. The number of defects in a reference book is the random variable $R$.

1. State the assumption of independence in context

The occurrence of one defect is independent of the occurrence of any other defects in the reference book.

2. State the assumption of constant average rate in context

The mean number of defects per reference book is constant.

1. Identify the distribution and formulate the required probability expression

From the given context and part (i), we model the number of defects $R$ as a Poisson distribution with mean 3, so $R \sim \text{Po}(3)$.
A book is donated if it has more than 7 defects, which corresponds to finding $\text{P}(R > 7)$.

2. Evaluate the probability using the complement rule

$\text{P}(R > 7) = 1 - \text{P}(R \le 7)$
Using the cumulative distribution function for the Poisson model (via calculator):
$\text{P}(R > 7) = 1 - 0.988095 = 0.011905$
$\text{P}(R > 7) \approx 0.0119 \text{ (to 3 s.f.)}$

1. Determine the distribution of the number of donated books out of 1000.

From part (ii), the probability that a reference book is donated away is $p = P(R > 7) = 1 - P(R \le 7) = 0.0119045$.
Let $Y$ be the number of reference books donated away out of 1000.
$Y \sim \text{B}(1000, 0.0119045)$

2. State and justify the appropriate approximation for the binomial distribution.

Since $n = 1000$ is large ($n \ge 50$) and $p = 0.0119045$ is small ($p < 0.1$), we approximate using a Poisson distribution with mean $\lambda = np$.
$\lambda = 1000 \times 0.0119045 = 11.9045$
Therefore, $Y \sim \text{Po}(11.9045)$ approximately.

3. Calculate the probability that at least 10 but no more than 15 books are donated away.

We require $P(10 \le Y \le 15)$.
$P(10 \le Y \le 15) = P(Y \le 15) - P(Y \le 9)$
$\approx 0.85117 - 0.25081$
$= 0.60036$
$\approx 0.600 \text{ (to 3 s.f.)}$

1. Define the random variable for the total number of defects and state its distribution.

Let $T = R + C$ be the total number of defects in the two books.
Since $R \sim \mathrm{Po}(3)$ and $C \sim \mathrm{Po}(5)$ are independent,
$T \sim \mathrm{Po}(3 + 5)$
$T \sim \mathrm{Po}(8)$

2. Calculate the probability that the total number of defects is fewer than 10.

We require $\mathrm{P}(T < 10)$. Since $T$ is discrete, this is equivalent to $\mathrm{P}(T \le 9)$.
Using a graphing calculator,
$\mathrm{P}(T \le 9) \approx 0.71662$
$\mathrm{P}(T \le 9) = 0.717 \text{ (to 3 s.f.)}$

1. Formulate an equation using the probability of at most 1 defect.

Let $C$ be the number of defects in a children's book. We have $C \sim \text{Po}(\lambda)$.
$\mathrm{P}(C \le 1) = 0.1$
$\mathrm{P}(C = 0) + \mathrm{P}(C = 1) = 0.1$
$e^{-\lambda} \frac{\lambda^0}{0!} + e^{-\lambda} \frac{\lambda^1}{1!} = 0.1$
$e^{-\lambda}(1 + \lambda) = 0.1$

2. Solve the equation for $\lambda$ using a graphing calculator.

By plotting $y_1 = e^{-x}(1 + x)$ and $y_2 = 0.1$ (or using an equation solver) on a graphing calculator:
$\lambda \approx 3.8897$
$\lambda = 3.89 \text{ (to 3 s.f.)}$