一家小公司生产高脚杯。每天，随机抽取 n 个高脚杯进行检查，并将发现有裂纹的高脚杯数量记为 X。 现在假设 X 服从分布 $B(n,p)$，其中 $n \ge 3$。

1. State the independence assumption in the context of the question

Whether a randomly chosen wine glass is cracked is independent of whether any other chosen wine glass is cracked.

2. State the constant probability assumption in the context of the question

The probability that a randomly chosen wine glass is cracked is constant for all wine glasses.

1. Set up equations for the mean and variance

For $X \sim B(n,p)$:
$\text{Mean} = np = 1.8$
$\text{Variance} = np(1-p) = 1.773$

2. Solve for $p$ by substituting the mean into the variance equation

$1.8(1-p) = 1.773$
$1-p = \frac{1.773}{1.8} = 0.985$
$p = 1 - 0.985 = 0.015$

3. Solve for $n$ using the value of $p$

$n(0.015) = 1.8$
$n = \frac{1.8}{0.015} = 120$

1. Set up the equation using the binomial probability formula.

$\mathrm{P}(X = 2) = 3\mathrm{P}(X = 3)$
$\binom{n}{2} p^2 (1-p)^{n-2} = 3 \binom{n}{3} p^3 (1-p)^{n-3}$

2. Expand the combinations and simplify the equation.

$\frac{n(n-1)}{2} p^2 (1-p)^{n-2} = 3 \left[ \frac{n(n-1)(n-2)}{6} \right] p^3 (1-p)^{n-3}$
Since $n \ge 3$ and assuming $0 < p < 1$, we can divide both sides by $\frac{n(n-1)}{2} p^2 (1-p)^{n-3}$:
$(1-p) = (n-2)p$

3. Solve for $p$ in terms of $n$.

$1 - p = np - 2p$
$1 = np - p$
$1 = p(n-1)$
$p = \frac{1}{n-1}$