Lesson Notes By Weeks and Term v5 - Grade 12
Counting and probability – Week 7 focus
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Counting and probability – Week 7 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: 7
Theme: General lesson support
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Counting and probability form a cornerstone of mathematical reasoning, enabling us to quantify uncertainty and make informed decisions. In South Africa, understanding probability is crucial in various fields, from analysing election results and predicting economic trends to assessing the risk of crime and evaluating the effectiveness of public health interventions. Being able to calculate probabilities allows individuals to critically evaluate information and make rational choices in everyday life, such as understanding lottery odds or assessing the fairness of insurance policies.
2.1 Tree Diagrams A tree diagram is a visual tool used to represent the possible outcomes of a sequence of events. Each branch of the tree represents a possible outcome, and the probabilities of each outcome are written along the branches.
Example 1: A bag contains 3 red balls and 2 blue balls. A ball is drawn at random, and then replaced. A second ball is drawn. Draw a tree diagram to represent this situation and calculate the probability of drawing two red balls.
Step 1: First Draw Red (R): Probability = 3/5 Blue (B): Probability = 2/5 Step 2: Second Draw (after replacement)
If the first ball was Red: Red (R): Probability = 3/5 Blue (B): Probability = 2/5 If the first ball was Blue: Red (R): Probability = 3/5 Blue (B): Probability = 2/5 Tree Diagram: ``` / R (3/5) / R (3/5)--- B (2/5) / Start \ B (2/5)--- R (3/5) \ \ B (2/5) ``` Probability of drawing two red balls: P(R and R) = P(R) P(R) = (3/5) * (3/5) = 9/25 Example 2 (Without Replacement): A bag contains 3 red balls and 2 blue balls. A ball is drawn at random, and not replaced. A second ball is drawn. Draw a tree diagram and calculate the probability of drawing two red balls.
Step 1: First Draw Red (R): Probability = 3/5 Blue (B): Probability = 2/5 Step 2: Second Draw (without replacement)
If the first ball was Red: Red (R): Probability = 2/4 (Since one red ball is already removed)
Blue (B): Probability = 2/4 If the first ball was Blue: Red (R): Probability = 3/4 Blue (B): Probability = 1/4 (Since one blue ball is already removed)
Tree Diagram: ``` / R (2/4) / R (3/5)--- B (2/4) / Start \ B (2/5)--- R (3/4) \ \ B (1/4) ``` Probability of drawing two red balls: P(R and R) = P(R) P(R|R) = (3/5) * (2/4) = 6/20 = 3/10 2.2 Venn Diagrams A Venn diagram is a visual representation of sets and their relationships. It uses overlapping circles to illustrate the common and distinct elements of different sets.
Example 3: In a class of 30 learners, 18 take Mathematics, 12 take Physical Sciences, and 5 take both Mathematics and Physical Sciences. Draw a Venn diagram to represent this situation and find the probability that a randomly selected learner takes either Mathematics or Physical Sciences.
Step 1: Draw the Venn Diagram Draw two overlapping circles. Label one circle "Mathematics (M)" and the other "Physical Sciences (P)". The overlapping region represents learners taking both subjects.
Step 2: Fill in the Venn Diagram The number of learners taking both subjects is 5, so write "5" in the overlapping region. The number of learners taking only Mathematics is 18 - 5 =
1
3. Write "13" in the Mathematics circle, outside the overlapping region. The number of learners taking only Physical Sciences is 12 - 5 =
7. Write "7" in the Physical Sciences circle, outside the overlapping region. The number of learners taking neither subject is 30 - (13 + 5 + 7) =
5. Write "5" outside both circles.
Step 3: Calculate the Probability The number of learners taking either Mathematics or Physical Sciences is 13 + 5 + 7 =
2
5. The probability that a randomly selected learner takes either Mathematics or Physical Sciences is P(M or P) = 25/30 = 5/
6. Formula Explanation: P(A or B) = P(A) + P(B) – P(A and B)
In this example: P(M or P) = P(M) + P(P) – P(M and P) = 18/30 + 12/30 - 5/30 = 25/30 = 5/6 2.3 Contingency Tables A contingency table (also called a two-way table) is a table that displays the frequency distribution of two or more categorical variables. It is a useful tool for analysing the relationship between these variables and calculating probabilities.
Example 4: A survey was conducted among 200 students at a university in Gauteng to investigate their preferred mode of transport to campus. The results are shown in the contingency table below: | Mode of Transport | Male | Female | Total | | ----------------- | ---- | ------ | ----- | | Car | 40 | 30 | 70 | | Bus | 30 | 50 | 80 | | Walk | 20 | 30 | 50 | | Total | 90 | 110 | 200 | Calculate the following probabilities: a) A randomly selected student is male. b) A randomly selected student uses the bus. c) A randomly selected student is male and uses the bus. d) A randomly selected student is male, given that they use the bus.
Solution: a) P(Male) = 90/200 = 9/20 b) P(Bus) = 80/200 = 2/5 c) P(Male and Bus) = 30/200 = 3/20 d) P(Male | Bus) = P(Male and Bus) / P(Bus) = (30/200) / (80/200) = 30/80 = 3/8 2.4 Mutually Exclusive Events Two events, A and B, are mutually exclusive if they cannot both occur at the same time. In other words, P(A and B) =
0. For mutually exclusive events, the probability of A or B occurring is: P(A or B) = P(A) + P(B)
Example 5: A standard six-sided die is rolled. What is the probability of rolling a 1 or a 6? Rolling a 1 and rolling a 6 are mutually exclusive events. P(1) = 1/6 P(6) = 1/6 P(1 or 6) = P(1) + P(6) = 1/6 + 1/6 = 2/6 = 1/3 2.5 Independent Events Two events, A and B, are independent if the occurrence of one does not affect the probability of the other.