To approach Part (iv), let’s break down what we need to find step by step.

Part (iv) Analysis

We are given that the MCC has at most one girl among the six selected students. We are tasked with finding the probability that, under this condition, the MCC contains exactly one girl.

This is a conditional probability problem. We’ll denote the events as follows:

• Let $A$ be the event that the MCC has at most one girl.
• Let $B$ be the event that the MCC has exactly one girl.

We want to find $P(B | A)$, which is the probability that the MCC contains exactly one girl, given that it contains at most one girl.

Using the formula for conditional probability: $P(B | A) = \frac{P(B \cap A)}{P(A)}$ Since $B$ is a subset of $A$ (if there’s exactly one girl, then we are indeed within the condition of having at most one girl), we can simplify this to: $P(B | A) = \frac{P(B)}{P(A)}$

Step 1: Calculate $P(A)$ (Probability of at most one girl)

To find $P(A)$, we consider the two scenarios that fulfill the condition of at most one girl:

1. No girls at all (all boys).
2. Exactly one girl and five boys.

1. Probability of Selecting All Boys

The probability of selecting all six students as boys can be calculated as follows:

• There are 20 students in total (15 boys and 5 girls).
• The probability of selecting 6 boys from a pool of 15 boys (and no girls) is a combination problem.

Thus, we can calculate the probability of all boys as: $P(\text{All boys}) = \frac{\binom{15}{6} \times \binom{5}{0}}{\binom{20}{6}}$

2. Probability of Selecting Exactly One Girl and Five Boys

The probability of selecting exactly one girl and five boys is: $P(\text{One girl, five boys}) = \frac{\binom{15}{5} \times \binom{5}{1}}{\binom{20}{6}}$

Then, the probability $P(A)$ (at most one girl) is: $P(A) = P(\text{All boys}) + P(\text{One girl, five boys})$

Step 2: Calculate $P(B)$ (Probability of Exactly One Girl)

We already calculated this as part of $P(A)$. So: $P(B) = P(\text{One girl, five boys}) = \frac{\binom{15}{5} \times \binom{5}{1}}{\binom{20}{6}}$

Step 3: Calculate $P(B | A)$

Now, substitute the values of $P(B)$ and $P(A)$ into the conditional probability formula: $P(B | A) = \frac{P(B)}{P(A)} = \frac{\frac{\binom{15}{5} \times \binom{5}{1}}{\binom{20}{6}}}{\frac{\binom{15}{6} \times \binom{5}{0} + \binom{15}{5} \times \binom{5}{1}}{\binom{20}{6}}}$

Simplify the fractions, and this will give you the conditional probability that, given at most one girl, exactly one girl is selected. Let me know if you'd like further simplification or calculation of these values.

Would you like more details on any specific step, or have questions on other parts?

---

Here are some follow-up questions to deepen understanding:

1. How do you calculate combinations like $\binom{15}{6}$?
2. What if there were more than 20 students in the class?
3. How would the probabilities change if the East class had a different gender ratio?
4. What is the difference between conditional and unconditional probability?
5. How do you interpret the result of $P(B | A)$ in practical terms?

Tip: When dealing with probabilities of multiple events, always break down complex conditions into smaller, manageable parts.