某工厂生产大量各种颜色的筹码，并将它们装入盒子中。每个盒子包含 36 个随机选择的筹码。平均而言，有 30% 的筹码是黄色的。你可以假设一个盒子中黄色筹码的数量服从二项分布。 75 个随机选择的盒子被装入一个纸箱中用于运输。

1. Define the random variable and state its probability distribution.

Let $X$ be the number of yellow chips in a box.
$X \sim \mathrm{B}(36, 0.3)$

2. Formulate the conditional probability expression based on the given conditions.

We need the probability that the box contains more than 4 yellow chips, given that it has at most 9 yellow chips.
$\mathrm{P}(X > 4 \mid X \le 9) = \frac{\mathrm{P}(4 < X \le 9)}{\mathrm{P}(X \le 9)}$

3. Evaluate the probability using binomial cumulative probabilities.

$\mathrm{P}(X > 4 \mid X \le 9) = \frac{\mathrm{P}(X \le 9) - \mathrm{P}(X \le 4)}{\mathrm{P}(X \le 9)}$
Using the graphing calculator:
$\mathrm{P}(X \le 9) \approx 0.31296$
$\mathrm{P}(X \le 4) \approx 0.010157$
$\mathrm{P}(X > 4 \mid X \le 9) = \frac{0.31296 - 0.010157}{0.31296} \approx 0.96755$
$\mathrm{P}(X > 4 \mid X \le 9) = 0.968 \text{ (to 3 s.f.)}$

1. Calculate the probability that a single randomly chosen box contains at most 9 yellow chips.

Let $X$ be the number of yellow chips in a box.
$X \sim \text{B}(36, 0.3)$

$P(X \le 9) \approx 0.29744$

2. Formulate the probability that the 60th box chosen is the 14th box with at most 9 yellow chips.

For the 60th box to be the 14th such box, there must be exactly 13 boxes with at most 9 yellow chips among the first 59 boxes, AND the 60th box must also have at most 9 yellow chips.

Let $Y$ be the number of boxes with at most 9 yellow chips in the first 59 boxes. Then $Y \sim \text{B}(59, 0.29744)$.

$\text{Required probability} = P(Y = 13) \times P(X \le 9)$
$= \binom{59}{13} (0.29744)^{13} (1 - 0.29744)^{46} \times 0.29744$

3. Evaluate the final probability.

$= \binom{59}{13} (0.29744)^{14} (0.70256)^{46}$
$= 0.0017859 \approx 0.00179 \text{ (to 3 s.f.)}$

1. Recall or calculate the probability that a randomly chosen box has at most 9 yellow chips.

Let $X$ be the number of yellow chips in a box.
$X \sim \mathrm{B}(36, 0.3)$
From earlier calculation, $p = \mathrm{P}(X \le 9) = 0.321806$.

2. Define the distribution for the number of such boxes in a single carton and find its mean and variance.

Let $Y$ be the number of boxes containing at most 9 yellow chips in a carton of 75 boxes.
$Y \sim \mathrm{B}(75, 0.321806)$

$\mathrm{E}(Y) = 75 \times 0.321806 = 24.1355$
$\mathrm{Var}(Y) = 75(0.321806)(1 - 0.321806) = 16.3686$

3. Apply the Central Limit Theorem to find the approximate distribution of the sample mean.

Let $\bar{Y}$ be the sample mean number of such boxes per carton for a sample of $n=30$ cartons.
Since $n = 30 \ge 30$, by the Central Limit Theorem, $\bar{Y}$ is approximately normally distributed:
$\bar{Y} \sim \mathrm{N}\left(24.1355, \frac{16.3686}{30}\right)$
$\bar{Y} \sim \mathrm{N}(24.1355, 0.54562)$

4. Calculate the probability that the mean number of boxes is more than 25.

$\mathrm{P}(\bar{Y} > 25) \approx 0.12092$
$\mathrm{P}(\bar{Y} > 25) = 0.121 \text{ (to 3 s.f.)}$