https://file.aiquickdraw.com/mathbot/file/1728421142801anhl9t82.jpg

Let's analyze the problem step by step.

### Given Data:
- Total number of students: 120
- Haskayne School of Business students (HS): 54
- 1st-Year students (1st): 71
- 1st-Year students who are **not** in the Haskayne School: 20

From the above, we can conclude the number of 1st-Year students in the Haskayne School:
\[
\text{HS} \cap \text{1st} = 71 - 20 = 51
\]

We can now use this information to complete the table, find the required probabilities, and answer the questions.

---

### Part (a) - Complete the Table:
We need to fill in the joint and marginal probabilities. First, let's calculate the number of students in each category:

1. **HS and 1st (HS ∩ 1st)**: 51
2. **HS and not 1st (HS ∩ 1stᶜ)**: 54 - 51 = 3
3. **Not HS and 1st (HSᶜ ∩ 1st)**: 20
4. **Not HS and not 1st (HSᶜ ∩ 1stᶜ)**: 120 - (51 + 3 + 20) = 46

Using these, we can calculate the probabilities by dividing by the total number of students, 120:

|              | HS              | HSᶜ             | Row Probabilities |
|--------------|-----------------|-----------------|-------------------|
| **1st**      | \( \frac{51}{120} = 0.4250 \) | \( \frac{20}{120} = 0.1667 \) | \( 0.4250 + 0.1667 = 0.5917 \) |
| **1stᶜ**     | \( \frac{3}{120} = 0.0250 \)  | \( \frac{46}{120} = 0.3833 \) | \( 0.0250 + 0.3833 = 0.4083 \) |
| **Column Probabilities** | \( 0.4250 + 0.0250 = 0.4500 \) | \( 0.1667 + 0.3833 = 0.5500 \) |  |

### Part (b) - Probability the student is either Haskayne or 1st-Year:
The event "Haskayne or 1st-Year" is the union of the two events:
\[
P(HS \cup 1st) = P(HS) + P(1st) - P(HS \cap 1st)
\]
\[
P(HS \cup 1st) = 0.4500 + 0.5917 - 0.4250 = 0.6167
\]

### Part (c) - Probability the student is neither Haskayne nor 1st-Year:
This event corresponds to the complement of "Haskayne or 1st-Year":
\[
P(\text{neither HS nor 1st}) = 1 - P(HS \cup 1st) = 1 - 0.6167 = 0.3833
\]

### Part (d) - Proportion of 1st-Year students who are not in the Haskayne School:
This is simply the probability of being in the intersection of HSᶜ and 1st:
\[
P(HSᶜ \cap 1st) = 0.1667
\]

### Part (e) - Are \( HS \) and \( 1st \) mutually exclusive?
Mutually exclusive events have no overlap, i.e., \( P(HS \cap 1st) = 0 \). However, in this case:
\[
P(HS \cap 1st) = 0.4250 \neq 0
\]
Thus, **HS** and **1st** are **not** mutually exclusive. The correct option is **B**.

### Part (f) - Are \( HS \) and \( 1st \) independent?
Two events are independent if:
\[
P(HS \cap 1st) = P(HS) \times P(1st)
\]
Let's check:
\[
P(HS) \times P(1st) = 0.4500 \times 0.5917 = 0.2663
\]
Since:
\[
P(HS \cap 1st) = 0.4250 \neq 0.2663
\]
Thus, **HS** and **1st** are **not** independent. The correct option is **D**.

---

Would you like further details or clarifications on any of these solutions?

Here are 5 related questions to deepen your understanding:
1. How would the table change if the total number of students increased to 150?
2. What is the probability of randomly selecting a student who is either in 1st-Year or not in the Haskayne School?
3. If a student is in the 1st year, what is the probability that they are also in the Haskayne School?
4. How would independence between \(HS\) and \(1st\) be determined using conditional probabilities?
5. What proportion of Haskayne students are in their 1st year?

**Tip**: When events are independent, the occurrence of one does not influence the probability of the other.

Consider your Statistics 213 lecture section. There are 120 students registered. Of these, 54 are Haskayne School of Business students (the remaining students belong to some other faculty). Moreover, 71 students are classified as 1st-Year students and 20 are 1st-Year students and who are not Haskayne students.

Analyze probability concepts for a group of 120 students, including Haskayne School of Business students and 1st-year students. Learn how to calculate probabilities, fill in joint probability tables, and understand mutually exclusive and independent events. Suitable for university-level students.

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