Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Calculate probability of combined events (A and B, A or B).
• Distinguish between mutually exclusive events.
• Use Venn diagrams to solve problems.
• Use tree diagrams to solve problems.
• Apply these principles to real-world problems.

Lesson summary

Probability is a fundamental concept in mathematics that deals with the likelihood of an event occurring. In Grade 10, we build upon the basic probability concepts learned in previous grades and delve into more complex scenarios, including combined events, mutually exclusive events, and the use of tree diagrams and Venn diagrams to represent probability. Understanding probability is crucial not only for academic success but also for making informed decisions in everyday life. From understanding weather forecasts to evaluating the risks associated with investments, probability plays a significant role in our lives, especially in a diverse and dynamic country like South Africa.

Lesson notes

2.1 Combined Events (A and B, A or B) When dealing with combined events, we are interested in the probability of two or more events occurring together (intersection) or the probability of at least one of the events occurring (union). "A and B" (Intersection): The probability of both event A and event B occurring. The notation for this is P(A ∩ B). "A or B" (Union): The probability of event A or event B occurring, or both. The notation for this is P(A ∪ B). The general formula for the probability of A or B is: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) The reason for subtracting P(A ∩ B) is to avoid double-counting the outcomes that are common to both A and B. 2.2 Mutually Exclusive Events Two events are mutually exclusive if they cannot occur at the same time. In other words, they have no outcomes in common. If A and B are mutually exclusive, then P(A ∩ B) =

0. For mutually exclusive events, the formula for the probability of A or B simplifies to: P(A ∪ B) = P(A) + P(B) 2.3 Venn Diagrams Venn diagrams are useful for visually representing the relationships between events and their probabilities. A rectangle represents the sample space (all possible outcomes), and circles within the rectangle represent individual events. The overlapping area between circles represents the intersection (A and B).

Example 1: Using a Venn Diagram In a class of 30 learners, 18 play soccer, 12 play netball, and 5 play both soccer and netball. Draw a Venn diagram to represent this information and find the probability that a learner chosen at random plays: (a) Soccer only (b) Netball only (c) Soccer or netball (d)

Neither soccer nor netball Solution: Draw a rectangle representing the sample space (30 learners). Draw two overlapping circles representing "Soccer" and "Netball". The overlapping area (A ∩ B) represents learners who play both, which is

5. The number of learners who play only soccer is 18 - 5 =

1

3. The number of learners who play only netball is 12 - 5 =

7. The number of learners who play neither is 30 - (13 + 7 + 5) =

5. Venn Diagram: ``` _______________________ | | | Soccer | | _________ | | | | | | | 13 | 5 | Netball | |_________|_________| | | | 7 | | | |_________| | | | | 5 (Neither) | |______________________| ``` (a) P(Soccer only) = 13/30 (b) P(Netball only) = 7/30 (c) P(Soccer or Netball) = (13 + 7 + 5) / 30 = 25/30 = 5/6 (d) P(Neither Soccer nor Netball) = 5/30 = 1/6 2.4 Tree Diagrams Tree diagrams are used to represent sequential events, where the outcome of one event affects the probability of the next event. Each branch of the tree represents a possible outcome, and the probabilities are written along the branches. To find the probability of a sequence of events, multiply the probabilities along the corresponding branches.

Example 2: Using a Tree Diagram A bag contains 3 red balls and 2 blue balls. A ball is drawn at random, its color is noted, and then it is replaced. Then, a second ball is drawn. Draw a tree diagram to represent this situation and find the probability of: (a) Drawing two red balls (b) Drawing a red ball followed by a blue ball (c) Drawing two balls of different colors Solution: First Draw: P(Red) = 3/5 P(Blue) = 2/5 Second Draw (after replacement): Since the ball is replaced, the probabilities for the second draw are the same as the first draw.

If the first ball was Red: P(Red) = 3/5 P(Blue) = 2/5 If the first ball was Blue: P(Red) = 3/5 P(Blue) = 2/5 Tree Diagram: ``` / Red (3/5) --- / Red (3/5) --> Red, Red / \ Blue (2/5) --> Red, Blue / Start - \ \ Blue (2/5) --- / Red (3/5) --> Blue, Red \ \ Blue (2/5) --> Blue, Blue ``` (a) P(Red, Red) = (3/5) * (3/5) = 9/25 (b) P(Red, Blue) = (3/5) * (2/5) = 6/25 (c) P(Different Colors) = P(Red, Blue) + P(Blue, Red) = (6/25) + (2/5)*(3/5) = 6/25 + 6/25 = 12/25 Example 3: Non-Mutually Exclusive Events Consider a standard deck of 52 playing cards. What is the probability of drawing a heart or a king?

Solution: Let H be the event of drawing a heart and K be the event of drawing a king. P(H) = 13/52 (There are 13 hearts in a deck) P(K) = 4/52 (There are 4 kings in a deck) P(H ∩ K) = 1/52 (There is one card that is both a heart and a king: the King of Hearts) Using the formula for the probability of A or B: P(H ∪ K) = P(H) + P(K) - P(H ∩ K) = (13/52) + (4/52) - (1/52) = 16/52 = 4/13 Therefore, the probability of drawing a heart or a king is 4/

1

3. Guided Practice (With Solutions)

Question 1: A die is rolled and a coin is tossed. What is the probability of getting a 4 on the die and a head on the coin?

Solution: Let A be the event of rolling a 4 on the die, and B be the event of getting a head on the coin. These events are independent. P(A) = 1/6 (There is one 4 on a six-sided die) P(B) = 1/2 (There is one head on a two-sided coin) Since the events are independent, P(A and B) = P(A) P(B) = (1/6) (1/2) = 1/12 Question 2: In a group of 40 learners, 22 take Mathematics, 18 take Physics, and 10 take both Mathematics and Physics.