There are ‘N’ total students in a class of which ‘m’ are female and the rest are male. If 5

students are picked at random for the “dress code enforcement committee” on

campus, calculate the probability that

a. The committee will only have female students.

b. The committee will only have male students.

Let's solve the problem by first understanding the setup and then applying the principles of combinatorics and probability.

### Given:
- Total number of students in the class: \( N \)
- Number of female students: \( m \)
- Number of male students: \( N - m \) (since the rest are male)
- Number of students selected for the committee: 5

### Part (a): Probability that the committee will only have female students

#### Step 1: Determine the total number of ways to select 5 students from \( N \) students
The number of ways to choose 5 students out of \( N \) is given by the combination formula:

\[
\text{Total combinations} = \binom{N}{5} = \frac{N!}{5!(N-5)!}
\]

#### Step 2: Determine the number of ways to select 5 female students from \( m \) female students
The number of ways to choose 5 female students out of \( m \) is given by:

\[
\text{Female combinations} = \binom{m}{5} = \frac{m!}{5!(m-5)!}
\]

#### Step 3: Calculate the probability
The probability that all 5 students selected are female is:

\[
P(\text{only females}) = \frac{\text{Female combinations}}{\text{Total combinations}} = \frac{\binom{m}{5}}{\binom{N}{5}} = \frac{\frac{m!}{5!(m-5)!}}{\frac{N!}{5!(N-5)!}} = \frac{m(m-1)(m-2)(m-3)(m-4)}{N(N-1)(N-2)(N-3)(N-4)}
\]

### Part (b): Probability that the committee will only have male students

#### Step 1: Determine the number of ways to select 5 male students from \( N - m \) male students
The number of ways to choose 5 male students out of \( N - m \) is given by:

\[
\text{Male combinations} = \binom{N-m}{5} = \frac{(N-m)!}{5!(N-m-5)!}
\]

#### Step 2: Calculate the probability
The probability that all 5 students selected are male is:

\[
P(\text{only males}) = \frac{\text{Male combinations}}{\text{Total combinations}} = \frac{\binom{N-m}{5}}{\binom{N}{5}} = \frac{\frac{(N-m)!}{5!(N-m-5)!}}{\frac{N!}{5!(N-5)!}} = \frac{(N-m)(N-m-1)(N-m-2)(N-m-3)(N-m-4)}{N(N-1)(N-2)(N-3)(N-4)}
\]

### Final Answers:
- **(a) The probability that the committee will only have female students is**:

\[
\boxed{P(\text{only females}) = \frac{\binom{m}{5}}{\binom{N}{5}}}
\]

- **(b) The probability that the committee will only have male students is**:

\[
\boxed{P(\text{only males}) = \frac{\binom{N-m}{5}}{\binom{N}{5}}}
\]

Do you want more details or have any questions?

Here are some related questions:

1. How does the probability change if the committee size is increased to 6?
2. What is the probability that the committee will have exactly 3 female students?
3. How would the probability change if \( m \) and \( N-m \) are very close in value?
4. What is the probability that the committee will have at least one female student?
5. How would you calculate the expected number of female students in the committee?

**Tip:** When dealing with probabilities in combinatorial problems, always remember to clearly identify the total possible outcomes and the favorable outcomes before calculating the probability.

Learn how to calculate the probability of forming a dress code enforcement committee composed entirely of female or male students from a class with specified numbers of female and male students. This problem involves combinatorics and probability theory, suitable for high school level students.

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