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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Okay, ngengnomhle, I'll solve Question 5 from the image you sent. Question 5.1: A group of 33 learners was surveyed at a school. The following information from the survey is given: • 2 learners play tennis, hockey and netball • 5 learners play hockey and netball • 7 learners play hockey and tennis • 6 learners play tennis and netball • A total of 18 learners play hockey • A total of 12 learners play tennis • 4 learners play netball ONLY 5.1.1 Determine the values of a, b, c, d and e. Step 1: Use the given information to fill in the Venn diagram regions. Let H be Hockey, N be Netball, T be Tennis. • N(H N T) = 2 (This is the center region, already labeled 2). • N(H T) = 7. From the diagram, N(H T) = a + 2. So, a + 2 = 7 a = 5. • N(H N) = 5. From the diagram, N(H N) = 3 + 2 = 5. (This matches the diagram). • N(T N) = 6. From the diagram, N(T N) = 4 + 2 = 6. (This matches the diagram). • N(Netball ONLY) = b = 4. • N(H) = 18. From the diagram, N(H) = c + a + 2 + 3. Substitute a=5: c + 5 + 2 + 3 = 18 c + 10 = 18 c = 8. • N(T) = 12. From the diagram, N(T) = d + a + 2 + 4. Substitute a=5: d + 5 + 2 + 4 = 12 d + 11 = 12 d = 1. Step 2: Calculate the value of e. The total number of learners surveyed is 33. The sum of all learners in the circles is: c + a + d + 3 + 2 + 4 + b = 8 + 5 + 1 + 3 + 2 + 4 + 4 = 27. The number of learners who do not play any sport (e) is the total learners minus those who play at least one sport. e = 33 - 27 = 6. The values are: a=5, b=4, c=8, d=1, e=6. 5.1.2 How many of these learners do not play any of the sports on the survey (that is netball, tennis or hockey)? Step 1: Identify the region representing learners who play none of the sports. This is the value e calculated in 5.1.1. The number of learners who do not play any of the sports is 6. 5.1.3 Write down the probability that a learner selected at random from this sample plays netball ONLY. Step 1: Identify the number of learners who play netball ONLY. From 5.1.1, b = 4 learners play netball ONLY. The total number of learners is 33. Step 2: Calculate the probability. P(Netball ONLY) = Number of learners playing netball ONLYTotal number of learners P(Netball ONLY) = (4)/(33) The probability that a learner plays netball ONLY is (4)/(33). 5.1.4 Determine the probability that a learner selected at random from this sample plays hockey or netball. Step 1: Identify the number of learners who play hockey or netball. This includes all regions within the Hockey (H) circle or the Netball (N) circle. N(H N) = c + a + 2 + 3 + b + 4 N(H N) = 8 + 5 + 2 + 3 + 4 + 4 = 26. The total number of learners is 33. Step 2: Calculate the probability. P(Hockey or Netball) = Number of learners playing hockey or netballTotal number of learners P(Hockey or Netball) = (26)/(33) The probability that a learner plays hockey or netball is (26)/(33). Question 5.2: In all South African schools, EVERY learner must choose to do either Mathematics or Mathematical Literacy. At a certain South African school, it is known that 60% of the learners are girls. The probability that a randomly chosen girl at the school does Mathematical Literacy is 55%. The probability that a randomly chosen boy at the school does Mathematical Literacy is 65%. Determine the probability that a learner selected at random from this school does Mathematics. Step 1: Define the probabilities given. Let G be the event a learner is a girl, B be the event a learner is a boy. Let ML be the event a learner does Mathematical Literacy, M be the event a learner does Mathematics. P(G) = 0.60 P(B) = 1 - P(G) = 1 - 0.60 = 0.40 P(ML | G) = 0.55 P(ML | B) = 0.65 Step 2: Calculate the probabilities of doing Mathematics for girls and boys. Since learners must choose either Mathematics or Mathematical Literacy: P(M | G) = 1 - P(ML | G) = 1 - 0.55 = 0.45 P(M | B) = 1 - P(ML | B) = 1 - 0.65 = 0.35 Step 3: Use the Law of Total Probability to find the overall probability of a learner doing Mathematics. P(M) = P(M | G) · P(G) + P(M | B) · P(B) P(M) = (0.45)(0.60) + (0.35)(0.40) P(M) = 0.27 + 0.14 P(M) = 0.41 The probability that a learner selected at random does Mathematics is 0.41. Question 5.3: William writes a Mathematics examination and an Accounting examination. He estimates that he has a 40% chance of passing the Mathematics examination. He estimates that he has a 60% chance of passing the Accounting examination. He estimates that he has a 30% chance of passing both. Determine the probability that William will fail Mathematics and Accounting. Step 1: Define the probabilities given. Let M be the event William passes Mathematics, A be the event William passes Accounting. P(M) = 0.40 P(A) = 0.60 P(M A) = 0.30 (Probability of passing both) Step 2: Determine the probability of passing Mathematics OR Accounting. Using the Addition Rule for probabilities: P(M A) = P(M) + P(A) - P(M A) P(M A) = 0.40 + 0.60 - 0.30 P(M A) = 1.00 - 0.30 P(M A) = 0.70 Step 3: Determine the probability of failing Mathematics AND failing Accounting. This is the complement of passing Mathematics OR Accounting, i.e., P((M A)'). P(Fail M and Fail A) = 1 - P(M A) P(Fail M and Fail A) = 1 - 0.70 P(Fail M and Fail A) = 0.30 The probability that William will fail Mathematics and Accounting is 0.30. 3 done, 2 left today. You're making progress.