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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Step 1: Analyze the given information for Question 4.1. Total number of learners = 700. Let C be the event that a learner uses cricket nets. Let T be the event that a learner uses tennis courts. P(C) = 72\% = 0.72 P(T) = 48\% = 0.48 P(neither) = P(C' T') = 8\% = 0.08 From P(C' T') = 0.08, we know that the probability of using at least one facility is: P(C T) = 1 - P(C' T') = 1 - 0.08 = 0.92 We also use the formula for the union of two events: P(C T) = P(C) + P(T) - P(C T) 0.92 = 0.72 + 0.48 - P(C T) 0.92 = 1.20 - P(C T) P(C T) = 1.20 - 0.92 = 0.28 This is the probability of using both facilities. Step 2: Solve Question 4.1.1. Determine the number of learners who use at least one of the facilities. The probability of using at least one facility is P(C T) = 0.92. Number of learners = P(C T) × Total learners Number of learners = 0.92 × 700 Number of learners = 644 The number of learners who use at least one facility is 644. Step 3: Solve Question 4.1.2. Determine the probability that a learner uses exactly one facility. The probability of using exactly one facility is P(C only) + P(T only). P(C only) = P(C) - P(C T) = 0.72 - 0.28 = 0.44 P(T only) = P(T) - P(C T) = 0.48 - 0.28 = 0.20 P(exactly one facility) = 0.44 + 0.20 = 0.64 Alternatively, P(exactly one facility) = P(C T) - P(C T) = 0.92 - 0.28 = 0.64. The probability that a learner uses exactly one facility is 0.64. Step 4: Solve Question 4.1.3. Determine whether the events 'using the cricket nets' and 'using the tennis courts' are independent or not. Two events A and B are independent if P(A B) = P(A) × P(B). Here, we check if P(C T) = P(C) × P(T). We found P(C T) = 0.28. Now, calculate P(C) × P(T): P(C) × P(T) = 0.72 × 0.48 = 0.3456 Since 0.28 ≠ 0.3456, the events are not independent. The events are not independent. Step 5: Analyze the given data for Question 4.2. The table shows data from the NSFAS financial aid office. Total students = 10 730. | | Receiving financial aid (A) | Not receiving financial aid (A') | Total | |---|---|---|---| | First time recipients (F) | 4 222 | 3 898 | 8 120 | | Repeating recipients (R) | 1 879 | 731 | 2 610 | | Total | 6 101 | 4 629 | 10 730 | Step 6: Solve Question 4.2.1. Determine the probability that a student selected at random is a student receiving financial aid. This is P(A). P(A) = Number of students receiving financial aidTotal number of students P(A) = (6101)/(10730) P(A) ≈ 0.5686 The probability that a student is receiving financial aid is (6101)/(10730) or approximately 0.5686. Step 7: Solve Question 4.2.2. Determine the probability that a student selected at random is a repeating learner and is not receiving financial aid. This is P(R A'). From the table, the number of repeating recipients not receiving financial aid is 731. P(R A') = Number of repeating recipients not receiving financial aidTotal number of students P(R A') = (731)/(10730) P(R A') ≈ 0.0681 The probability that a student is a repeating learner and is not receiving financial aid is (731)/(10730) or approximately 0.0681. Step 8: Solve Question 4.2.3. Determine the probability that a student selected at random is a first-time learner and is receiving financial aid. This is P(F A). From the table, the number of first-time recipients receiving financial aid is 4222. P(F A) = Number of first-time recipients receiving financial aidTotal number of students P(F A) = (4222)/(10730) P(F A) ≈ 0.3935 The probability that a student is a first-time learner and is receiving financial aid is (4222)/(10730) or approximately 0.3935. That's 2 down. 3 left today — send the next one.