In a group of five men and four women, there are three pairs of male and female business partners and three
teachers, where no teacher is in a business partnership. One man and one woman are selected at random. Find
the probability that they are:
a). both teachers b). in a business partnership with each other
c). each in a business partnership but not with each other.

Let’s define the situation and break it down carefully:

- There are **5 men** and **4 women**.
- Among them, there are **3 male-female business partnerships**.
- There are also **3 teachers** (but none of the teachers are in a business partnership).

### a) Probability that both selected are teachers:
We know that no teacher is in a business partnership. So, the teachers must be the remaining individuals who are not part of the 3 business partnerships. 

Thus, the teachers are:
- 2 men (since 3 out of the 5 men are in business partnerships, the other 2 must be teachers).
- 1 woman (since 3 out of the 4 women are in business partnerships, the other 1 must be a teacher).

So, the total number of teachers is 3: 2 men and 1 woman.

Now, we randomly select one man and one woman. The probability that both are teachers is the probability of selecting one of the two male teachers and the one female teacher.

The probability:
\[
P(\text{both are teachers}) = \frac{\text{number of male teachers}}{\text{total number of men}} \times \frac{\text{number of female teachers}}{\text{total number of women}} = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10}
\]

Thus, the probability that both are teachers is \( \frac{1}{10} \).

---

### b) Probability that they are in a business partnership with each other:
We are selecting one man and one woman. There are 3 business partnerships, each consisting of 1 man and 1 woman.

Thus, the probability that the selected pair is one of these 3 business partnerships is the ratio of favorable outcomes (choosing one of the 3 partnerships) to total possible outcomes (choosing any man and any woman).

The total number of ways to choose one man and one woman is \( 5 \times 4 = 20 \).

The probability:
\[
P(\text{in a business partnership with each other}) = \frac{3}{20}
\]

Thus, the probability that they are in a business partnership with each other is \( \frac{3}{20} \).

---

### c) Probability that each is in a business partnership but not with each other:
This time, we want to select one man and one woman, both of whom are in business partnerships, but **not** with each other. 

There are 3 men in partnerships and 3 women in partnerships. If we select one of these men, we have 2 remaining women who are in partnerships but not with him (since we want them to not be in the same partnership).

So, the total number of favorable outcomes is \( 3 \times 2 = 6 \) (since for each of the 3 partnered men, there are 2 women in different partnerships).

The total number of possible outcomes is still \( 5 \times 4 = 20 \).

The probability:
\[
P(\text{each in a business partnership but not with each other}) = \frac{6}{20} = \frac{3}{10}
\]

Thus, the probability that each is in a business partnership but not with each other is \( \frac{3}{10} \).

---

### Summary of probabilities:
a) Probability that both are teachers: \( \frac{1}{10} \)  
b) Probability that they are in a business partnership with each other: \( \frac{3}{20} \)  
c) Probability that each is in a business partnership but not with each other: \( \frac{3}{10} \)

---

Would you like further clarification on any part of this? Here are some related questions:

1. How would the probabilities change if there were more business partners?
2. What if the group contained an equal number of men and women?
3. How would you calculate the probability of selecting a teacher and a non-teacher at random?
4. What if we had the number of men and women in partnerships reversed?
5. How could the calculation change if teachers were allowed to have partnerships?

**Tip:** In probability problems, it's useful to break things down into clear subsets and count favorable outcomes relative to total outcomes.

In a group of five men and four women, there are three pairs of male and female business partners and three teachers, where no teacher is in a business partnership. One man and one woman are selected at random. Find the probability that they are: a) both teachers, b) in a business partnership with each other, c) each in a business partnership but not with each other.

This problem involves calculating the probability of selecting teachers and business partners from a group of five men and four women. Using probability concepts, we find the chances of picking both teachers, a business partnership, or each partner not matched together. Perfect for Grades 9-11 students learning probability and combinatorics.

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