A company produces ceramic vases. A fixed number of randomly chosen vases are inspected each day and the number of defective vases found on a day is denoted by $X$.

1. State the constant probability assumption in context

The probability that any given vase is defective must be constant for all vases inspected.

2. State the independence assumption in context

The event that a vase is defective must be independent of whether any other vase is defective.

1. Express the required probability using the complement rule

$\mathrm{P}(X > 2) = 1 - \mathrm{P}(X \leqslant 2)$

2. Evaluate the probability using a graphing calculator

$= 1 - 0.883108\dots \\ \approx 0.117 \text{ (to 3 s.f.)}$

1. Calculate the probability of finding more than 2 defective vases on a single day (from part (b)).

$X \sim \text{B}(30, 0.04)$
$P(X > 2) = 1 - P(X \le 2) \approx 0.11706$

2. Define a new binomial random variable for the 5-day week.

Let $Y$ be the number of days in a 5-day working week where more than 2 defective vases are found.
$Y \sim \text{B}(5, 0.11706)$

3. Calculate the probability that this event happens on at most 1 day.

We require $P(Y \le 1)$.
$P(Y \le 1) = P(Y = 0) + P(Y = 1)$
$P(Y \le 1) = (1 - 0.11706)^5 + \binom{5}{1}(0.11706)(1 - 0.11706)^4$
$P(Y \le 1) \approx 0.892$

1. Formulate the condition as the probability of finding exactly 5 defective vases in a 5-day week.

Let $Y$ be the total number of defective vases in a 5-day working week.
Since 30 vases are inspected each day over 5 days, a total of 150 vases are inspected.
Thus, $Y \sim \mathrm{B}(150, 0.04)$.
The probability of finding exactly 5 defective vases in the week is:
$P(Y = 5) = \binom{150}{5}(0.04)^5(0.96)^{145} = 591600036(0.04)^5(0.96)^{145}$

2. Determine the number of ways to distribute 5 defective vases exactly across 2 consecutive days.

There are 4 pairs of consecutive days in a 5-day week (e.g., Days 1-2, 2-3, 3-4, 4-5).
For any specific pair of 2 days, 60 vases are inspected. The number of ways to choose 5 defective vases from these 60 is $\binom{60}{5}$.
To ensure the 5 defective vases are found on exactly 2 days (not just 1), we subtract the cases where all 5 vases fall entirely on the first day or entirely on the second day:
$\text{Ways for a specific pair} = \binom{60}{5} - 2\binom{30}{5}$
$= 5461512 - 2(142506) = 5176500$
(Alternatively, summing the possible daily splits $(1,4), (2,3), (3,2), (4,1)$ gives $\binom{30}{1}\binom{30}{4} + \binom{30}{2}\binom{30}{3} + \binom{30}{3}\binom{30}{2} + \binom{30}{4}\binom{30}{1} = 5176500$)

3. Calculate the joint probability of finding exactly 5 defective vases such that they all fall on exactly 2 consecutive days.

Since there are 4 such mutually exclusive pairs of consecutive days, the total number of ways this specific arrangement can occur is:
$4 \times 5176500 = 20706000$
The probability of this intersection event occurring over the 5 days (where the 5 defectives happen exactly as configured, and the remaining 3 days yield 0 defectives) is:
$P(\text{exactly 2 consecutive days} \cap Y=5) = 20706000(0.04)^5(0.96)^{145}$

4. Compute the conditional probability.

Using the conditional probability formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$:
$P(\text{exactly 2 consecutive days} \mid Y=5) = \frac{20706000(0.04)^5(0.96)^{145}}{591600036(0.04)^5(0.96)^{145}}$
$= \frac{20706000}{591600036}$
$\approx 0.0350 \text{ (to 3 s.f.)}$