Lesson Notes By Weeks and Term v5 - Grade 11
Probability – Week 9 focus
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Probability – Week 9 focus
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Subject: Mathematics
Class: Grade 11
Term: 3rd Term
Week: 9
Theme: General lesson support
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Probability is the study of chance and uncertainty. It allows us to quantify the likelihood of events happening, from the simple toss of a coin to the complexities of weather forecasting or predicting election outcomes. Understanding probability is crucial in making informed decisions in various aspects of life. For South African learners, understanding probability can be applied to things like lottery odds (Powerball, Lotto), understanding insurance policies, predicting sports outcomes (like rugby or soccer results), assessing risks related to agriculture (likelihood of drought affecting crop yields), and even interpreting health statistics (like the prevalence of a disease in a community).
Conditional Probability: Conditional probability is the probability of an event occurring, given that another event has already occurred. It's denoted by P(A|B), which reads "the probability of A given B." The formula for calculating conditional probability is: P(A|B) = P(A ∩ B) / P(B), where P(B) ≠
0. Here: P(A|B) is the conditional probability of event A occurring given that event B has already occurred. P(A ∩ B) is the probability of both events A and B occurring (the intersection of A and B). P(B) is the probability of event B occurring.
Example 1: A survey of 100 students at a school in Soweto revealed the following: 60 students like soccer, 40 students like rugby, and 25 students like both soccer and rugby. If a student is randomly selected and it is known that they like soccer, what is the probability that they also like rugby?
Solution: Let S be the event that a student likes soccer, and R be the event that a student likes rugby.
We are given: P(S) = 60/100 = 0.6 P(R) = 40/100 = 0.4 P(S ∩ R) = 25/100 = 0.25 We want to find P(R|S), the probability that a student likes rugby given that they like soccer.
Using the formula: P(R|S) = P(R ∩ S) / P(S) = 0.25 / 0.6 = 5/12 ≈ 0.4167 Therefore, the probability that a student likes rugby, given that they like soccer, is approximately 41.67%.
Independence of Events: Two events A and B are independent if the occurrence of one event does not affect the probability of the other event occurring. The test for independence can be done using either of the following conditions: P(A ∩ B) = P(A) * P(B) P(A|B) = P(A), provided P(B) > 0 P(B|A) = P(B), provided P(A) > 0 If either of these conditions hold true, the events A and B are independent. Otherwise, they are dependent.
Example 2: A fair coin is tossed twice. Let A be the event that the first toss results in heads, and B be the event that the second toss results in tails. Are events A and B independent?
Solution: Since the coin is fair, P(Heads) = P(Tails) = 0.5 P(A) = 0.5 (Probability of getting heads on the first toss) P(B) = 0.5 (Probability of getting tails on the second toss) P(A ∩ B) = P(Heads on the first toss AND Tails on the second toss) = 0.5 0.5 = 0.25 Using the test for independence: P(A) P(B) = 0.5 0.5 = 0.25 Since P(A ∩ B) = P(A) * P(B), the events A and B are independent.
Tree Diagrams: Tree diagrams are useful for visualizing and calculating probabilities in situations where events occur in sequence. Each branch represents a possible outcome, and the probabilities are written along the branches.
Example 3: A bag contains 5 red balls and 3 blue balls. A ball is drawn at random from the bag, and then a second ball is drawn without replacement. Use a tree diagram to find the probability that both balls are red.
Solution: First Draw: Probability of drawing a red ball (R1) = 5/8 Probability of drawing a blue ball (B1) = 3/8 Second Draw (without replacement): If the first ball was red (R1), then there are now 4 red balls and 3 blue balls left, making a total of 7 balls. Probability of drawing a red ball (R2|R1) = 4/7 Probability of drawing a blue ball (B2|R1) = 3/7 If the first ball was blue (B1), then there are now 5 red balls and 2 blue balls left, making a total of 7 balls. Probability of drawing a red ball (R2|B1) = 5/7 Probability of drawing a blue ball (B2|B1) = 2/7 To find the probability that both balls are red, we follow the path R1 -> R2: P(R1 ∩ R2) = P(R1) P(R2|R1) = (5/8) (4/7) = 20/56 = 5/14 Therefore, the probability that both balls are red is 5/
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4. Two-Way Contingency Tables: Two-way contingency tables are used to organize data based on two categorical variables. They provide a clear way to visualize the relationship between these variables and to calculate probabilities.
Example 4: A group of 200 people were surveyed about their smoking habits.
The results are shown in the table below: | | Smoker | Non-Smoker | Total | |-----------------|--------|------------|-------| | Male | 40 | 60 | 100 | | Female | 25 | 75 | 100 | | Total | 65 | 135 | 200 | What is the probability that a randomly selected person is a smoker, given that the person is male?
Solution: We want to find P(Smoker | Male).
From the table: Number of male smokers = 40 Total number of males = 100 P(Smoker | Male) = (Number of male smokers) / (Total number of males) = 40/100 = 0.4 Therefore, the probability that a randomly selected person is a smoker, given that the person is male, is 0.
4. Inclusion-Exclusion Principle: The inclusion-exclusion principle states that for any two events A and B, the probability of either A or B occurring (or both) is given by: P(A∪B) = P(A) + P(B) – P(A∩B)
Example 5: In a class, 60% of the students like Mathematics, 40% like Science, and 20% like both Mathematics and Science. What is the probability that a randomly selected student likes either Mathematics or Science (or both)?
Solution: Let M be the event that a student likes Mathematics, and S be the event that a student likes Science.