Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

This page supports the lesson note with a companion video and a short classroom-ready summary.

For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.

Performance objectives

• Calculate probabilities of combined events ('A and B', 'A or B').
• Understand independent and dependent events.
• Calculate conditional probabilities.
• Use Venn diagrams, two-way tables, and tree diagrams.
• Evaluate probabilities related to everyday risks.

Lesson summary

This week, we delve into the fascinating world of probability, specifically focusing on combined events and how probability informs our understanding of everyday risk. Probability isn't just a theoretical concept; it's a powerful tool that helps us make informed decisions in various aspects of our lives, from understanding the likelihood of winning the Lotto to assessing the risks associated with different transportation options or even understanding the spread of diseases like HIV/AIDS in our communities.

Lesson notes

2.1 Combined Events: Combined events involve two or more events occurring together. We are typically interested in finding the probability of these events happening in conjunction. The two main types of combined events we'll focus on are: 'A and B' (Intersection): The event where both event A and event B occur. 'A or B' (Union): The event where either event A or event B (or both) occurs. 2.2 Independent vs.

Dependent Events: The relationship between events is crucial for calculating their combined probabilities.

Independent Events: Two events are independent if the outcome of one event does not affect the outcome of the other.

For independent events A and B: P(A and B) = P(A)

P(B)

Dependent Events: Two events are dependent if the outcome of one event does affect the outcome of the other. This leads to conditional probability. 2.3 Conditional Probability: Conditional probability refers to the probability of an event A occurring, given that another event B has already occurred. It is denoted as P(A|B) and read as "the probability of A given B." Formula: P(A|B) = P(A and B) / P(B), provided P(B) ≠ 0. 2.4 Mutually Exclusive Events: Two events are mutually exclusive (or disjoint) if they cannot both occur at the same time.

If A and B are mutually exclusive: P(A and B) = 0 P(A or B) = P(A) + P(B) 2.5 The Addition Rule: A general formula for calculating the probability of 'A or B': P(A or B) = P(A) + P(B) - P(A and B)

Note: If A and B are mutually exclusive, then P(A and B) = 0, and the formula simplifies to P(A or B) = P(A) + P(B). 2.6 Two-Way Tables and Venn Diagrams: These are visual tools that help organize information about events and their probabilities.

Two-Way Tables: Show the frequencies or probabilities of different combinations of events.

Venn Diagrams: Use overlapping circles to represent events, with the overlapping region representing the intersection ('and') and the entire area covered representing the union ('or'). 2.7 Tree Diagrams: Tree diagrams are useful for visualizing and calculating probabilities in sequential events. Each branch represents a possible outcome, and the probabilities are written along the branches. To find the probability of a specific path, multiply the probabilities along that path. 2.8 Everyday Risk: Many aspects of our lives involve risk. Understanding probability helps us quantify and compare these risks, leading to better decisions.

Examples include: Health Risks: Probability of contracting diseases, effectiveness of treatments.

Financial Risks: Probability of investment success/failure, insurance premiums.

Safety Risks: Probability of accidents, crime victimization.

Worked example

Example 1: Independent Events

A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, its colour is noted, and it is replaced. Then, a second ball is drawn. What is the probability that both balls are red?

Solution:

Event A: First ball is red. P(A) = 5/8

Event B: Second ball is red. P(B) = 5/8 (Since the first ball was replaced, the events are independent).

P(A and B) = P(A) P(B) = (5/8) * (5/8) = 25/64

Explanation: Since the ball is replaced, the outcome of the first draw doesn't influence the second.

Therefore, we simply multiply the probabilities of each independent event.

Example 2: Dependent Events (Conditional Probability)

A box contains 10 light bulbs, 3 of which are defective. Two bulbs are selected at random, without replacement. What is the probability that both bulbs are defective?

Solution:

Event A: First bulb is defective. P(A) = 3/10

Event B: Second bulb is defective, given the first one was defective. P(B|A) = 2/9 (Since one defective bulb was removed, there are only 2 defective bulbs left out of 9 total bulbs).

P(A and B) = P(A) P(B|A) = (3/10) * (2/9) = 6/90 = 1/15