一位水果商从一家拥有大型种植园的供应商处采购橙子，该种植园中 1% 的橙子是腐烂的。该供应商在采摘橙子时不考虑橙子是否腐烂，并随机地将 20 个橙子装入每个箱子中。

1. State the assumption regarding the independence of events in context.

Whether an orange is rotten is independent of whether any other orange in the box is rotten.

2. State the assumption regarding constant probability in context.

The probability of any randomly chosen orange being rotten is constant at $0.01$.

1. Identify the parameters of the binomial distribution for $X$.

From the problem context, there are $20$ oranges in a box and the probability of an orange being rotten is $1\% = 0.01$.
$X \sim \mathrm{B}(20, 0.01)$

2. Calculate the probability that a box contains between 2 and 5 rotten oranges inclusive.

We need to find $\mathrm{P}(2 \le X \le 5)$.

Using a graphing calculator:
$\begin{aligned} \mathrm{P}(2 \le X \le 5) &= \mathrm{P}(X \le 5) - \mathrm{P}(X \le 1) \\ &= 0.016858\dots \\ &\approx 0.0169 \text{ (to 3 s.f.)} \end{aligned}$

1. Calculate the expected value of the binomial distribution using the formula $\mathrm{E}(X) = np$

$\mathrm{E}(X) = 20 \times 0.01 = 0.2$

1. Calculate the probability that a single randomly chosen box is substandard (from part b).

From part (b), we first find the probability that a box is substandard. The random variable $X$ denotes the number of rotten oranges in a box of 20.
$X \sim \text{B}(20, 0.01)$

$\text{P}(\text{box is substandard}) = \text{P}(2 \le X \le 5)$
Using a graphing calculator:
$\text{P}(2 \le X \le 5) = \text{P}(X \le 5) - \text{P}(X \le 1)$
$\phantom{\text{P}(\text{box is substandard})} \approx 0.016859$

2. Define a new binomial random variable for the truckload of 10 boxes and calculate the probability that not more than 2 are substandard.

Let $Y$ be the number of substandard boxes in a truckload of 10.
$Y \sim \text{B}(10, 0.016859)$

We need to find the probability that not more than 2 boxes are substandard, which is $\text{P}(Y \le 2)$.
Using a graphing calculator:
$\text{P}(Y \le 2) \approx 0.99967$
$\phantom{\text{P}(Y \le 2)} = 1.00 \text{ (to 3 s.f.)}$

1. Identify the mutually exclusive cases that result in rejecting the truckload.

Based on the fruit seller's conditions, the truckload is rejected in two mutually exclusive cases:

• Case 1: The first box opened has $>1$ rotten oranges. ($X_1 > 1$)
• Case 2: The first box opened has exactly 1 rotten orange AND the second box opened has $>1$ rotten oranges. ($X_1 = 1$ and $X_2 > 1$)

2. Calculate the necessary binomial probabilities for a single box.

From earlier parts, the number of rotten oranges in a box follows $X \sim \mathrm{B}(20, 0.01)$.
$\mathrm{P}(X = 1) = \binom{20}{1}(0.01)^1(0.99)^{19} \approx 0.16523$
$\mathrm{P}(X > 1) = 1 - \mathrm{P}(X = 0) - \mathrm{P}(X = 1)$
$\mathrm{P}(X > 1) = 1 - (0.99)^{20} - 0.16523 \approx 0.016859$

3. Calculate the total probability of rejecting the truckload.

Since the boxes are independent, we use the addition and multiplication principles:
$\mathrm{P}(\text{Reject}) = \mathrm{P}(X_1 > 1) + [\mathrm{P}(X_1 = 1) \times \mathrm{P}(X_2 > 1)]$
$\mathrm{P}(\text{Reject}) = 0.016859 + (0.16523 \times 0.016859)$
$\mathrm{P}(\text{Reject}) \approx 0.016859 + 0.0027856$
$\mathrm{P}(\text{Reject}) \approx 0.019645$
$\mathrm{P}(\text{Reject}) = 0.0196 \text{ (to 3 s.f.)}$