A bakery produces batches of cookies. On average, the proportion of flawed cookies produced is $p$, where $0 < p < 1$. The cookies are packed in boxes of 20. The number of flawed cookies in a box of cookies is denoted by $C$.

1. State one assumption for the binomial model in the context of the problem.

Assumption: Whether a cookie is flawed is independent of whether any other cookie in the box is flawed.

(Alternatively: The probability $p$ of a cookie being flawed is constant for all cookies in the box.)

1. Formulate the equation for $p$ using the binomial probability formula.

$C \sim \text{B}(20, p)$
Given $P(C = 0 \text{ or } 1) = 0.15$,
$\binom{20}{0}p^0(1-p)^{20} + \binom{20}{1}p^1(1-p)^{19} = 0.15$
$(1-p)^{20} + 20p(1-p)^{19} = 0.15$

2. Solve the resulting equation numerically using a graphing calculator.

Using a graphing calculator to solve the equation $(1-p)^{20} + 20p(1-p)^{19} - 0.15 = 0$ for $0 < p < 1$,
$p \approx 0.15901$
$p = 0.159 \text{ (to 3 s.f.)}$

1. Find the probability that a single box of cookies is rejected.

Let $C$ be the number of flawed cookies in a box.
$C \sim \text{B}(20, 0.08)$

A box is rejected if it contains 4 or more flawed cookies.
$P(\text{box rejected}) = P(C \ge 4)$
$P(C \ge 4) = 1 - P(C \le 3)$
Using a graphing calculator,
$P(C \ge 4) \approx 0.070615$

2. Define the binomial distribution for the number of rejected boxes.

Let $R$ be the number of rejected boxes out of 10 randomly chosen boxes.
$R \sim \text{B}(10, 0.070615)$

3. Calculate the probability of having between 2 and 5 rejected boxes inclusive.

We require $P(2 \le R \le 5)$.
$P(2 \le R \le 5) = P(R \le 5) - P(R \le 1)$

Using a graphing calculator,
$P(2 \le R \le 5) \approx 0.15384$
$\approx 0.154 \text{ (to 3 s.f.)}$

1. Set up the conditional probability relation based on the given events.

Let $B$ be the event that exactly $3$ boxes are rejected out of the sample of $15$.
Let $A$ be the event that the $3$rd rejected box is the $15$th box.
Since $A$ implies that there are exactly $3$ rejected boxes in total, $A \subseteq B$, which means $P(A \cap B) = P(A)$.
We need to find $P(A|B) = \frac{P(A)}{P(B)}$.

2. Formulate expressions for $P(A)$ and $P(B)$.

Let $p_r$ be the probability of a single box being rejected.
The probability of exactly $3$ rejected boxes out of $15$ is given by the binomial distribution:
$P(B) = \binom{15}{3} p_r^3 (1-p_r)^{12}$
For event $A$ to occur, there must be exactly $2$ rejected boxes in the first $14$ boxes, followed by a rejected $15$th box:
$P(A) = \left[ \binom{14}{2} p_r^2 (1-p_r)^{12} \right] \times p_r = \binom{14}{2} p_r^3 (1-p_r)^{12}$

3. Calculate the final conditional probability.

$P(A|B) = \frac{\binom{14}{2} p_r^3 (1-p_r)^{12}}{\binom{15}{3} p_r^3 (1-p_r)^{12}}$
$P(A|B) = \frac{\binom{14}{2}}{\binom{15}{3}} = \frac{91}{455} = \frac{1}{5}$