An eatery is giving out free burgers in a promotion to celebrate its anniversary. A total of 80 burgers, consisting of 8 different flavours, 10 from each flavour, are to be given out. The burgers are randomly chosen and given out one at a time. Of the 8 burger flavours, the 'Curry' and 'Spicy' burgers are most popular among customers. In the first four burgers to be given out, the events $A$ and $B$ are defined as follows: $\begin{aligned} A: & \text{ At least 2 'Curry' burgers.} \\ B: & \text{ Exactly one 'Spicy' burger.} \end{aligned}$

1. Find the number of ways event $B$ can occur.

Event $B$ is getting exactly 1 'Spicy' burger and 3 non-'Spicy' burgers in the first four burgers.
Number of ways to choose 1 'Spicy' out of 10 and 3 non-'Spicy' out of the remaining $70$:
$n(B) = \binom{10}{1} \binom{70}{3} = 10 \times 54740 = 547400$

2. Find the number of ways event $A \cap B$ can occur.

Event $A \cap B$ means exactly 1 'Spicy' burger and at least 2 'Curry' burgers. The 3 non-Spicy burgers must consist of either 2 'Curry' and 1 'Other', or 3 'Curry' and 0 'Other' (where 'Other' represents the $80 - 10 - 10 = 60$ remaining burgers).
$n(A \cap B) = \binom{10}{1} \left[ \binom{10}{2}\binom{60}{1} + \binom{10}{3}\binom{60}{0} \right]$
$= 10 \left[ 45(60) + 120(1) \right]$
$= 10(2700 + 120) = 28200$

3. Calculate the conditional probability $\text{P}(A \mid B)$.

$\text{P}(A \mid B) = \frac{\text{P}(A \cap B)}{\text{P}(B)} = \frac{n(A \cap B)}{n(B)}$
$= \frac{28200}{547400}$
$= \frac{141}{2737} \approx 0.0515 \text{ (to 3 s.f.)}$

1. State or calculate $\text{P}(A \mid B)$ from part (i).

$\text{From (i), given exactly one 'Spicy' burger is chosen, the other 3 burgers must be chosen from the remaining 70 (10 'Curry', 60 others).}$
$\text{To have at least 2 'Curry' burgers:}$
$\text{P}(A \mid B) = \frac{\binom{10}{2}\binom{60}{1} + \binom{10}{3}\binom{60}{0}}{\binom{70}{3}}$
$\text{P}(A \mid B) = \frac{2700 + 120}{54740} = \frac{2820}{54740} = \frac{141}{2737} \approx 0.0515$

2. Calculate the unconditional probability $\text{P}(A)$.

$\text{Calculate the unconditional probability } \text{P}(A)\text{:}$
$\text{P}(A) = \frac{\binom{10}{2}\binom{70}{2} + \binom{10}{3}\binom{70}{1} + \binom{10}{4}\binom{70}{0}}{\binom{80}{4}}$
$\text{P}(A) = \frac{108675 + 8400 + 210}{1581580} = \frac{117285}{1581580} = \frac{23457}{316316} \approx 0.0742$

3. Compare the probabilities to conclude on the independence of $A'$ and $B'$.

$\text{For } A \text{ and } B \text{ to be independent, } \text{P}(A \mid B) \text{ must equal } \text{P}(A).$
$\text{Since } \frac{141}{2737} \neq \frac{23457}{316316} \text{ (i.e. } 0.0515 \neq 0.0742\text{), } A \text{ and } B \text{ are not independent.}$
$\text{It is a standard probability property that events } A \text{ and } B \text{ are independent if and only if their complements } A' \text{ and } B' \text{ are independent.}$
$\text{Therefore, } A' \text{ and } B' \text{ are also not independent.}$