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
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 12
Term: 3rd Term
Week: 6
Theme: General lesson support
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This week, we delve into the fascinating world of counting techniques and probability. Understanding how to count possibilities and calculate probabilities is crucial not only for acing your Mathematics exams but also for making informed decisions in various aspects of life. From assessing the odds of winning the Lotto to evaluating risks in business ventures, these concepts are invaluable. In South Africa, where diverse cultures and situations often involve making choices based on likelihoods, a solid grasp of counting and probability empowers you to be a more analytical and informed citizen.
2. 1. Fundamental Counting Principle (FCP) 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 one thing, n ways to do another, p ways to do a third thing, and so on, then there are m x n x p x ... ways to do all of them.
Example 1: A South African restaurant offers 3 choices of starters, 5 choices of main courses, and 2 choices of desserts. How many different three-course meals are possible?
Solution: Using the FCP, the total number of meals is 3 x 5 x 2 = 30 different meals.
Why it works: Each starter can be combined with each main course, creating 3 x 5 = 15 combinations. Each of these 15 combinations can then be combined with each of the 2 desserts, resulting in 15 x 2 = 30 possible meals. 2.
2. Permutations A permutation is an arrangement of objects in a specific order. 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 calculated as: P(n, r) = n! / (n - r)! Where '!' denotes the factorial (e.g., 5! = 5 x 4 x 3 x 2 x 1).
Example 2: In a class of 20 students, a teacher wants to choose a head prefect, a deputy head prefect, and a secretary. How many different ways can these positions be filled?
Solution: This is a permutation problem because the order matters (head prefect is different from deputy). We have n = 20 and r =
3. P(20, 3) = 20! / (20 - 3)! = 20! / 17! = 20 x 19 x 18 = 6840 ways.
Why it works: The first position (head prefect) can be filled in 20 ways. Once the head prefect is chosen, the second position (deputy) can be filled in 19 ways (since one student is already head prefect). Finally, the third position (secretary) can be filled in 18 ways. Thus, there are 20 x 19 x 18 possible outcomes. 2.
3. Combinations A combination is a selection of objects without regard to order. The number of combinations of n distinct objects taken r at a time is denoted by C(n, r) or n C r or ( n r ) and is calculated as: C(n, r) = n! / (r! (n - r)!)
Example 3: A lottery requires selecting 6 numbers from a set of
4
9. How many different lottery tickets are possible?
Solution: This is a combination problem because the order in which you choose the numbers doesn't matter. We have n = 49 and r =
6. C(49, 6) = 49! / (6! (49 - 6)!) = 49! / (6! 43!) = (49 x 48 x 47 x 46 x 45 x 44) / (6 x 5 x 4 x 3 x 2 x 1) = 13,983,816 possible tickets.
Why it works: We are selecting a group of 6 from 49, and the order doesn't matter. The formula adjusts for overcounting that would occur if we treated different orderings of the same 6 numbers as distinct. 2.
4. Probability of an Event The probability of an event A, denoted P(A), is the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely. P(A) = (Number of favorable outcomes) / (Total number of possible outcomes) The probability of any event lies between 0 and 1 (inclusive). P(A) = 0 means the event is impossible, and P(A) = 1 means the event is certain.
Example 4: A bag contains 5 red balls and 3 blue balls. What is the probability of drawing a red ball at random?
Solution: There are 5 red balls (favorable outcomes) and a total of 8 balls (possible outcomes). P(Red) = 5/8 Why it works: We directly apply the definition of probability. The probability of drawing a red ball is the proportion of red balls in the bag. 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. If A and B are independent, then P(A and B) = P(A) x P(B).
Dependent Events: Two events are dependent if the occurrence of one does affect the probability of the other. If A and B are dependent, then P(A and B) = P(A) x P(B|A), where P(B|A) is the conditional probability of B given that A has occurred.
Example 5: Consider a standard deck of 52 cards.
Independent: Drawing a card, replacing it, and then drawing another card. The outcome of the first draw does not affect the probabilities of the second draw.
Dependent: Drawing a card and not replacing it, and then drawing another card. The probabilities for the second draw are affected because there are only 51 cards left.
Example 6: In a class of 30 students, 12 are taking mathematics and 15 are taking accounting. If 5 students are taking both, what is the probability that a randomly chosen student is taking mathematics given that they are taking accounting?
Solution: P(Mathematics | Accounting) = P(Mathematics and Accounting) / P(Accounting) P(Mathematics and Accounting) = 5/30 P(Accounting) = 15/30 P(Mathematics | Accounting) = (5/30) / (15/30) = 5/15 = 1/3 Why it works: The conditional probability formula narrows the sample space to only those students taking accounting, and then calculates the proportion of those students who are also taking mathematics. 2.6.