Question 4.2.1: Represent the information about students' post-graduation plans in a Venn diagram.

The number of students with plans is the total number of students minus those with no plans:

$$N(W \cup B) = N(U) - N(W \cup B)'$$ $$N(W \cup B) = 530 - 60$$ $$N(W \cup B) = 470$$

Using the formula for the union of two sets:

$$N(W \cup B) = N(W) + N(B) - N(W \cap B)$$ $$470 = 340 + 180 - N(W \cap B)$$ $$470 = 520 - N(W \cap B)$$ $$N(W \cap B) = 520 - 470$$ $$N(W \cap B) = 50$$

Now we can answer the specific questions.

Question 4.2.1

Represent the above information in a Venn diagram.

Based on the calculations:

• Number of students in both W and B: $N(W \cap B) = 50$
• Number of students only in W: $N(W) - N(W \cap B) = 340 - 50 = 290$
• Number of students only in B: $N(B) - N(W \cap B) = 180 - 50 = 130$
• Number of students with no plans (outside W and B): $N(W \cup B)' = 60$

A Venn diagram would show two overlapping circles, W and B, within a rectangle representing the universal set U.
The intersection (W $\cap$ B) would contain 50.
The part of W not overlapping with B would contain 290.
The part of B not overlapping with W would contain 130.
The area outside both circles but inside the rectangle would contain 60.

Question 4.2.2

Determine the number of students that are planning to utilise their NCV knowledge and skills in both the working environment AND by opening their businesses.

This is the number of students in the intersection of W and B, which was calculated in Step 2.

$$N(W \cap B) = 50$$

The final answer is $\boxed{50}$.

Question 4.2.3

Determine the probability that if a student is chosen randomly in College ABC, the student will be the one that has a plan about where he/she will utilise his/her NCV skills.

This is the probability of a student having any plan, which is $P(W \cup B)$.

$$P(W \cup B) = \frac{N(W \cup B)}{N(U)}$$

From Step 1, $N(W \cup B) = 470$.

$$P(W \cup B) = \frac{470}{530}$$ $$P(W \cup B) = \frac{47}{53}$$

The final answer is $\boxed*{\frac{47}{53}}$.

Question 4.2.4

Are the events utilising knowledge and skills in the working environment (W) and utilising knowledge and skills by opening one's business (B) INDEPENDENT Events? Substantiate your answer.

For two events W and B to be independent, the following condition must be met:

$$P(W \cap B) = P(W) \times P(B)$$

$$P(W) = \frac{N(W)}{N(U)} = \frac{340}{530} = \frac{34}{53}$$ $$P(B) = \frac{N(B)}{N(U)} = \frac{180}{530} = \frac{18}{53}$$ $$P(W \cap B) = \frac{N(W \cap B)}{N(U)} = \frac{50}{530} = \frac{5}{53}$$

Calculate $P(W) \times P(B)$:

$$P(W) \times P(B) = \frac{34}{53} \times \frac{18}{53}$$ $$P(W) \times P(B) = \frac{34 \times 18}{53 \times 53}$$ $$P(W) \times P(B) = \frac{612}{2809}$$

Compare this to $P(W \cap B)$:

$$P(W \cap B) = \frac{5}{53}$$

To compare, we can convert $\frac{5}{53}$ to have a denominator of 2809:

$$\frac{5}{53} = \frac{5 \times 53}{53 \times 53} = \frac{265}{2809}$$

Since $\frac{265}{2809} \neq \frac{612}{2809}$, the condition for independence is not met.
Therefore, the events W and B are NOT independent.

The final answer is $\boxed{\text{No, the events are NOT independent. This is because } P(W \cap B) = \frac{5}{53} \text{ while } P(W) \times P(B) = \frac{34}{53} \times \frac{18}{53} = \frac{612}{2809}. \text{ Since } \frac{5}{53} \neq \frac{612}{2809}, \text{ the events are not independent.}}$.