This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here's the solution for Question 4.2. First, let's define the events and given numbers: Total number of students, N(U) = 530. Number of students planning to utilise knowledge in the working environment, N(W) = 340. Number of students planning to open their own businesses, N(B) = 180. Number of students with no plans, N(W B)' = 60. Step 1: Calculate the number of students with at least one plan. The number of students with plans is the total number of students minus those with no plans: N(W B) = N(U) - N(W B)' N(W B) = 530 - 60 N(W B) = 470 Step 2: Calculate the number of students with both plans (intersection). Using the formula for the union of two sets: N(W B) = N(W) + N(B) - N(W B) 470 = 340 + 180 - N(W B) 470 = 520 - N(W B) N(W B) = 520 - 470 N(W 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 B) = 50 Number of students only in W: N(W) - N(W B) = 340 - 50 = 290 Number of students only in B: N(B) - N(W B) = 180 - 50 = 130 Number of students with no plans (outside W and B): N(W B)' = 60 A Venn diagram would show two overlapping circles, W and B, within a rectangle representing the universal set U. The intersection (W 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 B) = 50 The final answer is 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 B). P(W B) = (N(W B))/(N(U)) From Step 1, N(W B) = 470. P(W B) = (470)/(530) P(W B) = (47)/(53) The final answer is *(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 B) = P(W) × P(B) Step 1: Calculate the individual probabilities. P(W) = (N(W))/(N(U)) = (340)/(530) = (34)/(53) P(B) = (N(B))/(N(U)) = (180)/(530) = (18)/(53) P(W B) = (N(W B))/(N(U)) = (50)/(530) = (5)/(53) Step 2: Check the independence condition. Calculate P(W) × P(B): P(W) × P(B) = (34)/(53) × (18)/(53) P(W) × P(B) = (34 × 18)/(53 × 53) P(W) × P(B) = (612)/(2809) Compare this to P(W B): P(W B) = (5)/(53) To compare, we can convert (5)/(53) to have a denominator of 2809: (5)/(53) = (5 × 53)/(53 × 53) = (265)/(2809) Since (265)/(2809) ≠ (612)/(2809), the condition for independence is not met. Therefore, the events W and B are NOT independent. The final answer is No, the events are NOT independent. This is because P(W B) = (5)/(53) while P(W) × P(B) = (34)/(53) × (18)/(53) = (612)/(2809). Since (5)/(53) ≠ (612)/(2809), the events are not independent.. 3 done, 2 left today. You're making progress.