Lesson Notes By Weeks and Term v5 - Grade 12
Counting and probability – Week 6 focus
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Counting and probability – Week 6 focus
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Subject: Mathematics
Class: Grade 12
Term: 3rd Term
Week: 6
Theme: General lesson support
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Counting and Probability are fundamental mathematical tools that allow us to understand and predict the likelihood of events occurring. This week, we will delve deeper into these concepts, building upon your Grade 11 knowledge and introducing more complex scenarios. These skills are not just theoretical; they have practical applications in various aspects of South African life, from understanding lottery odds to analysing crime statistics, assessing risk in financial investments, and even predicting election outcomes. Imagine designing a survey to gauge public opinion on a local issue - a solid understanding of probability is crucial for accurately interpreting the results.
2. 1.
Fundamental Counting Principle: The fundamental counting principle states that if there are m ways to do one thing and n ways to do another, then there are m x n ways to do both. This extends to any number of independent events. The "and" usually indicates multiplication. Think of it like building a meal deal. If you have 3 choices for a main course and 4 choices for a drink, you have 3 * 4 = 12 possible meal deal combinations.
Example 1: A restaurant in Durban offers 5 different starters, 8 different main courses, and 3 different desserts. How many different three-course meals can a customer choose?
Solution: Number of starters = 5 Number of main courses = 8 Number of desserts = 3 Total number of meals = 5 x 8 x 3 = 120 2.
2. Permutations: A permutation is an arrangement of objects in a specific order. The order matters! The formula for the number of permutations of r objects chosen from n distinct objects is: P(n, r) = n! / (n - r)! Where n! (n factorial) = n x (n - 1) x (n - 2) x ... x 2 x 1 Example 2: How many different ways can you arrange 3 letters from the word "ORANGE"?
Solution: Here, n = 6 (total number of letters) and r = 3 (number of letters to arrange) P(6, 3) = 6! / (6 - 3)! = 6! / 3! = (6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1) = 6 x 5 x 4 = 120 There are 120 different arrangements of 3 letters from the word "ORANGE." 2.
3. Combinations: A combination is a selection of objects where the order does not matter. The formula for the number of combinations of r objects chosen from n distinct objects is: C(n, r) = n! / (r! * (n - r)!)
Example 3: A soccer team of 11 players is to be selected from a group of 16 players. How many different teams are possible?
Solution: Here, n = 16 (total number of players) and r = 11 (number of players to be selected). The order in which the players are selected does not matter, so we use combinations. C(16, 11) = 16! / (11! (16 - 11)!) = 16! / (11! 5!) = (16 x 15 x 14 x 13 x 12) / (5 x 4 x 3 x 2 x 1) = 4368 There are 4368 different possible teams. 2.
4. Permutations with Identical Objects: When dealing with permutations where some objects are identical, we need to adjust the formula to avoid overcounting. If there are n objects in total, with n 1 of one kind, n 2 of another kind, and so on, then the number of distinct permutations is: n! / (n 1 ! n 2 ! ... * n k !)
Example 4: How many different ways can you arrange the letters in the word "BANANA"?
Solution: Here, n = 6 (total number of letters) Number of A's (n 1 ) = 3 Number of N's (n 2 ) = 2 Number of B's (n 3 ) = 1 Number of arrangements = 6! / (3! 2! 1!) = (6 x 5 x 4 x 3 x 2 x 1) / ((3 x 2 x 1) x (2 x 1) x 1) = 720 / (6 x 2 x 1) = 720 / 12 = 60 There are 60 different arrangements of the letters in "BANANA." 2.
5. Independent and Dependent Events: Independent Events: Two events A and B are independent if the occurrence of one does not affect the probability of the other. P(A and B) = P(A) x P(B)
Dependent Events: Two events A and B are dependent if the occurrence of one does affect the probability of the other. P(A and B) = P(A) x P(B|A) (where P(B|A) is the probability of B given that A has already occurred).
Example 5: A bag contains 5 red marbles and 3 blue marbles. A marble is drawn, its colour noted, and then replaced. A second marble is drawn. What is the probability that both marbles are red?
Solution: Since the first marble is replaced, the events are independent. P(Red on first draw) = 5/8 P(Red on second draw) = 5/8 P(Red and Red) = (5/8) x (5/8) = 25/64 Example 6: A bag contains 5 red marbles and 3 blue marbles. A marble is drawn, its colour noted, and not replaced. A second marble is drawn. What is the probability that both marbles are red?
Solution: Since the first marble is not replaced, the events are dependent. P(Red on first draw) = 5/8 P(Red on second draw | Red on first draw) = 4/7 (because there are now only 4 red marbles and 7 total marbles left) P(Red and Red) = (5/8) x (4/7) = 20/56 = 5/14 2.
6. Conditional Probability: The probability of event A occurring given that event B has already occurred is denoted by P(A|B) and is calculated as: P(A|B) = P(A ∩ B) / P(B) Where P(A ∩ B) is the probability of both A and B occurring.
Example 7: In a class, 60% of the students play soccer, and 40% play rugby. 25% of the students play both soccer and rugby. What is the probability that a student plays rugby given that they play soccer?
Solution: Let S = event that a student plays soccer Let R = event that a student plays rugby P(S) = 0.6 P(R) = 0.4 P(S ∩ R) = 0.25 P(R|S) = P(R ∩ S) / P(S) = 0.25 / 0.6 = 5/12 2.
7. Venn Diagrams in Probability: Venn diagrams are visual tools used to represent the relationships between sets (events). They are particularly helpful for understanding and calculating probabilities involving unions, intersections, and complements of events.
Example 8: A survey was conducted among 100 students.