A small company makes wine glasses. Each day, n randomly chosen wine glasses are checked and the number of wine glasses found to be cracked is denoted by X. Assume now that X has the distribution $B(n,p)$, where $n \ge 3$.

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 $n$.

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