Lesson Notes By Weeks and Term v5 - Grade 12
Counting and probability – Week 8 focus
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Counting and probability – Week 8 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 12
Term: 3rd Term
Week: 8
Theme: General lesson support
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Counting and probability are fundamental concepts in mathematics with wide-ranging applications in everyday life. From understanding the chances of winning the Lotto to assessing the risks involved in financial investments, the ability to calculate probabilities is crucial for making informed decisions. In South Africa, where socioeconomic challenges often require strategic planning and risk assessment, a solid understanding of counting and probability is particularly valuable. For example, understanding probability can assist in making informed decisions related to insurance, budgeting, and even predicting election outcomes.
2.1 The Fundamental Counting Principle The Fundamental Counting Principle (FCP) 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 principle extends to any number of events. If there are m ways to do the first thing, n ways to do the second thing, p ways to do the third thing, and so on, then there are m x n x p x … ways to do all the things.
Example 1: A restaurant in Durban offers 3 choices of starters, 5 choices of main courses, and 2 choices of desserts. How many different three-course meals are possible?
Solution: Applying the FCP, the number of possible meals is 3 x 5 x 2 = 30. 2.2 Permutations A permutation is an arrangement of objects in a specific order. The order matters in permutations. The number of permutations of n distinct objects taken r at a time is denoted by P(n, r) or n P r , and is given by: P(n, r) = n! / (n - r)! where n! (n factorial) is the product of all positive integers less than or equal to n.
Example 2: How many different 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5, without repetition?
Solution: We are choosing 3 digits out of 5, and the order matters (e.g., 123 is different from 321). This is a permutation problem. P(5, 3) = 5! / (5 - 3)! = 5! / 2! = (5 x 4 x 3 x 2 x 1) / (2 x 1) =
6
0. There are 60 different 3-digit numbers. 2.3 Combinations A combination is a selection of objects where the order does not matter. The number of combinations of n distinct objects taken r at a time is denoted by C(n, r) or n C r , and is given by: C(n, r) = n! / (r! * (n - r)!)
Example 3: A committee of 4 students is to be chosen from a group of 10 students. How many different committees can be formed?
Solution: We are choosing 4 students out of 10, and the order in which they are chosen does not matter. This is a combination problem. C(10, 4) = 10! / (4! (10 - 4)!) = 10! / (4! 6!) = (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1) =
2
1
0. There are 210 different committees. 2.4 Probability Using Combinations and Permutations Probability is the measure of the likelihood that an event will occur. It is calculated as the number of favorable outcomes divided by the total number of possible outcomes. We can often use combinations and permutations to calculate the number of favorable outcomes and the total number of possible outcomes.
Example 4: A bag contains 5 red balls and 3 blue balls. If 2 balls are drawn at random, what is the probability that both balls are red?
Solution: Total number of ways to choose 2 balls from 8: C(8, 2) = 8! / (2! 6!) = 28 Number of ways to choose 2 red balls from 5: C(5, 2) = 5! / (2! 3!) = 10 Probability of choosing 2 red balls: P(2 red) = 10 / 28 = 5 / 14 2.5 Probability Rules Addition Rule: For any two events A and B, P(A or B) = P(A) + P(B) - P(A and B). If A and B are mutually exclusive (cannot occur at the same time), then P(A and B) = 0, so P(A or B) = P(A) + P(B).
Product Rule: For independent events A and B, P(A and B) = P(A) x P(B).
Example 5: A die is rolled. What is the probability of rolling a 2 or a 4?
Solution: Rolling a 2 and rolling a 4 are mutually exclusive events. P(2) = 1/6, P(4) = 1/
6. P(2 or 4) = P(2) + P(4) = 1/6 + 1/6 = 1/3. 2.6 Dependent and Independent Events Independent Events: Two events are independent if the outcome of one event does not affect the outcome of the other event.
Dependent Events: Two events are dependent if the outcome of one event does affect the outcome of the other event.
Example 6: A card is drawn from a standard deck of 52 cards. The card is not replaced, and a second card is drawn. What is the probability that both cards are aces?
Solution: This is a dependent event because the first draw affects the number of cards remaining and the number of aces remaining. P(Ace on first draw) = 4/52 = 1/13 P(Ace on second draw | Ace on first draw) = 3/51 = 1/17 P(Ace on first draw and Ace on second draw) = (1/13) x (1/17) = 1/221 Guided Practice (With Solutions)
Question 1: How many different ways can the letters of the word "SUCCESS" be arranged?
Solution: The word "SUCCESS" has 7 letters, with S occurring 3 times, C occurring 2 times, and U and E each occurring once. The formula for permutations with repetitions is n! / (n1! n2! … * nk!), where n is the total number of objects and n1, n2, …, nk are the number of repetitions of each distinct object.
Therefore, the number of arrangements is 7! / (3! 2! 1! * 1!) = (7 x 6 x 5 x 4 x 3 x 2 x 1) / ( (3 x 2 x 1) x (2 x 1) x 1 x 1) =
4
2
0. Commentary: This question requires applying the concept of permutations with repetitions. Students often forget to divide by the factorials of the repetition counts, leading to an incorrect answer.
Question 2: A soccer team of 11 players needs to be selected from a squad of 15 players. How many different teams can be formed?
Solution: This is a combination problem because the order in which the players are selected does not matter. C(15, 11) = 15! / (11!