Lesson Notes By Weeks and Term v5 - Grade 10
Probability – Week 3 focus
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Probability – Week 3 focus
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Subject: Mathematics
Class: Grade 10
Term: Term 4
Week: 3
Theme: General lesson support
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Probability is a fundamental concept in mathematics that deals with the likelihood of an event occurring. In this week, we will build upon the basic probability concepts introduced earlier and delve into more complex scenarios involving combined events and conditional probability. Understanding probability is crucial because it helps us make informed decisions in everyday life, from assessing risks to understanding statistical data presented in the news and in business. Think about lotteries, weather forecasts, or even predicting the outcome of a soccer match - all rely on probability.
2.1 Combined Events: Often, we're interested in the probability of more than one event happening. We need tools like Venn diagrams and tree diagrams to visualize and calculate these probabilities effectively.
Venn Diagrams: These diagrams visually represent sets and their relationships. They are particularly useful for calculating the probability of events occurring together (intersection) or either event occurring (union). Intersection (A ∩ B): The event where both A and B occur. The probability is denoted as P(A ∩ B). Union (A ∪ B): The event where either A or B (or both) occur. The probability is denoted as P(A ∪ B).
Formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Tree Diagrams: These diagrams are useful for representing sequential events, especially when the outcome of one event affects the probability of the next. Each branch represents a possible outcome, and the probability is written along the branch. To find the probability of a sequence of events, multiply the probabilities along the corresponding branches.
Example 1: Using a Venn Diagram In a class of 30 students, 18 take Mathematics, 12 take Science, and 5 take both. What is the probability that a randomly selected student takes either Mathematics or Science or both?
Solution: Let M = event that a student takes Mathematics, and S = event that a student takes Science. P(M) = 18/30 P(S) = 12/30 P(M ∩ S) = 5/30 Using the formula: P(M ∪ S) = P(M) + P(S) - P(M ∩ S) P(M ∪ S) = (18/30) + (12/30) - (5/30) = 25/30 = 5/6 Therefore, the probability that a randomly selected student takes either Mathematics or Science or both is 5/
6. Example 2: Using a Tree Diagram A bag contains 3 red balls and 2 blue balls. A ball is drawn at random and its colour noted, and then returned to the bag. A second ball is then drawn. What is the probability of drawing a red ball, then a blue ball?
Solution: Probability of drawing a red ball first (R1) = 3/5 Probability of drawing a blue ball second (B2), given that the red ball was returned = 2/5 The probability of drawing a red ball then a blue ball: P(R1 and B2) = P(R1) P(B2) = (3/5) (2/5) = 6/25 Therefore, the probability is 6/25. 2.2 Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur at the same time. In other words, if one event happens, the other cannot. For example, flipping a coin cannot result in both heads and tails simultaneously.
Key Property: If A and B are mutually exclusive, then P(A ∩ B) =
0. Formula: P(A ∪ B) = P(A) + P(B) (for mutually exclusive events)
Example 3: Mutually Exclusive Events What is the probability of rolling either a 2 or a 5 on a standard six-sided die?
Solution: Let A = event of rolling a 2, and B = event of rolling a
5. These are mutually exclusive. P(A) = 1/6 P(B) = 1/6 P(A ∪ B) = P(A) + P(B) = (1/6) + (1/6) = 2/6 = 1/3 Therefore, the probability is 1/3. 2.3 Conditional Probability: Conditional probability is the probability of an event occurring, given that another event has already occurred. It's denoted as P(A|B), which reads "the probability of A given B." Formula: P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0 Example 4: Conditional Probability A survey of 100 learners at a high school revealed that 60 learners take Maths, 40 take Science and 20 take both Maths and Science. A learner is selected at random. Determine the probability that the learner takes Science given that they take Maths.
Solution: Let M = event that a learner takes Maths, and S = event that a learner takes Science. P(M) = 60/100 = 0.6 P(S) = 40/100 = 0.4 P(M ∩ S) = 20/100 = 0.2 Using the formula: P(S|M) = P(S ∩ M) / P(M) P(S|M) = 0.2 / 0.6 = 1/3 Therefore, the probability is 1/3. 2.4 Independent and Dependent Events: Independent Events: Two events are independent if the occurrence of one does not affect the probability of the other.
Key Property: If A and B are independent, then P(A ∩ B) = P(A)
P(B)
Dependent Events: Two events are dependent if the occurrence of one affects the probability of the other. Conditional probability deals with dependent events.
Example 5: Independent Events A coin is flipped and a die is rolled. What is the probability of getting tails on the coin and a 4 on the die?
Solution: Let T = event of getting tails on the coin, and F = event of rolling a 4 on the die. P(T) = 1/2 P(F) = 1/6 Since these events are independent: P(T ∩ F) = P(T) P(F) = (1/2) (1/6) = 1/12 Therefore, the probability is 1/
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2. Guided Practice (With Solutions)
Question 1: A survey was conducted among 80 learners about their favourite sport. 40 learners like soccer, 30 like rugby, and 10 like both. What is the probability that a randomly selected learner likes either soccer or rugby (or both)?
Solution: Let S = event that a learner likes soccer, and R = event that a learner likes rugby. P(S) = 40/80 P(R) = 30/80 P(S ∩ R) = 10/80 P(S ∪ R) = P(S) + P(R) - P(S ∩ R) = (40/80) + (30/80) - (10/80) = 60/80 = 3/4 Therefore, the probability is 3/4.