**Scenario:**

Shaldon High School has two AS mathematics classes, East and West. Both classes have 15 boys and 5 girls.

The East class selects 3 students randomly to represent it on the MCC (Mathematics Consultative Committee).

**Questions:**

**Part (i):**

Complete the probability tree diagram (Fig. 2) to show all possible choices of boys and girls from the East class.

**Part (ii):**

Find the probability that the East class is represented by:

(A) Three boys

(B) Two boys and one girl

**Part (iii):**

The West class also chooses 3 students randomly to represent it on the MCC, making a total of 6 students on the committee.

Find the probability that the MCC contains:

(A) One girl and five boys

(B) At most one girl

**Part (iv):**

Given that the MCC contains at most one girl, find the probability that it contains exactly one girl.
help with part 4

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.

Shaldon High School has two AS mathematics classes, East and West. Both classes have 15 boys and 5 girls. The East class selects 3 students randomly to represent it on the MCC (Mathematics Consultative Committee).

Questions:

**Part (i):** Complete the probability tree diagram (Fig. 2) to show all possible choices of boys and girls from the East class.

**Part (ii):** Find the probability that the East class is represented by:

(A) Three boys

(B) Two boys and one girl

**Part (iii):** The West class also chooses 3 students randomly to represent it on the MCC, making a total of 6 students on the committee.

Find the probability that the MCC contains:

(A) One girl and five boys

(B) At most one girl

**Part (iv):** Given that the MCC contains at most one girl, find the probability that it contains exactly one girl.

Explore probability questions involving the selection of students from two AS mathematics classes with 15 boys and 5 girls. Includes calculations for probabilities of specific gender combinations, use of conditional probability, and combination formulas. Suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation