Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Calculate the probability of combined events using diagrams and tables.
• Differentiate between independent and dependent events.
• Analyze probabilities associated with everyday risks.
• Use probability to make informed decisions.

Lesson summary

Probability is all about understanding how likely something is to happen. This is crucial in many areas of life, from making informed financial decisions to understanding health risks. In South Africa, where socio-economic factors can heavily influence risk, understanding probability is not just academic; it’s a vital life skill. This week, we'll focus on combined events (what happens when we look at multiple events together) and how probability plays out in everyday risks we face. From understanding the likelihood of winning the Lotto to evaluating the risk of contracting HIV/AIDS, these skills empower you to make informed decisions and protect yourselves.

Lesson notes

2.1 Combined Events Combined events are situations where we're interested in the probability of two or more events occurring. There are several ways events can be combined, leading to different methods for calculating probabilities.

Union of Events (OR): The union of two events, A and B, denoted A ∪ B, represents the event that either A or B (or both) occur.

Intersection of Events (AND): The intersection of two events, A and B, denoted A ∩ B, represents the event that both A and B occur. 2.2 Calculating Probabilities of Combined Events General Addition Rule: P(A ∪ B) = P(A) + P(B) – P(A ∩ B) This formula is essential because it accounts for the possibility of overlap between events A and B. If A and B are mutually exclusive (they cannot both occur at the same time), then P(A ∩ B) = 0, and the formula simplifies to P(A ∪ B) = P(A) + P(B).

Conditional Probability: P(A|B) represents the probability of event A occurring given that event B has already occurred.

The formula is: P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠

0. Independent Events: Two events, A and B, are independent if the occurrence of one does not affect the probability of the other.

For independent events: P(A ∩ B) = P(A) P(B). Also, P(A|B) = P(A) and P(B|A) = P(B).

Dependent Events: Events that are not independent are dependent. In this case, the occurrence of one event does affect the probability of the other. 2.3 Visualizing Probability: Venn Diagrams, Tree Diagrams, and Contingency Tables Venn Diagrams: These diagrams visually represent sets and their relationships, making it easy to see the overlap (intersection) between events. The universal set represents all possible outcomes.

Tree Diagrams: These are particularly useful for sequential events, where the outcome of one event affects the probabilities of subsequent 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.

Contingency Tables (Two-Way Tables): These tables display the frequencies of different combinations of two or more categorical variables. They are useful for calculating conditional probabilities. 2.4 Everyday Risk and Probability Probability is used to quantify risk in various real-life scenarios. Understanding the probabilities associated with these risks enables informed decision-making. This involves assessing probabilities related to health, finance, safety, and other aspects of life.

Example 1: Health Risk (HIV/AIDS) Let's say that in a South African community, 15% of the adult population is HIV positive. Also, a test for HIV has a 99% accuracy rate, meaning that if someone is HIV positive, the test will be positive 99% of the time (sensitivity).

However, it also has a 1% false positive rate, meaning that if someone is HIV negative, the test will still show positive 1% of the time (1-specificity). If someone tests positive, what is the probability that they actually have HIV? This is a conditional probability problem. We want P(HIV Positive | Test Positive). Let HIV+ = HIV Positive, HIV- = HIV Negative, Test+ = Test Positive, Test- = Test Negative. P(HIV+) = 0.15 P(HIV-) = 0.85 P(Test+|HIV+) = 0.99 P(Test-|HIV+) = 0.01 P(Test+|HIV-) = 0.01 P(Test-|HIV-) = 0.99 We can use Bayes' Theorem or calculate directly. P(Test+) = P(Test+|HIV+) P(HIV+) + P(Test+|HIV-) P(HIV-) P(Test+) = (0.99 0.15) + (0.01 0.85) = 0.1485 + 0.0085 = 0.157 P(HIV+|Test+) = P(Test+|HIV+) * P(HIV+) / P(Test+) P(HIV+|Test+) = (0.99 * 0.15) / 0.157 = 0.1485 / 0.157 ≈ 0.946 Therefore, even though the test is 99% accurate, the probability that someone who tests positive actually has HIV is about 94.6%. This highlights the importance of understanding conditional probability and prevalence rates.

Example 2: Financial Risk (Lotto) The South African Lotto requires you to choose 6 numbers from a pool of 1 to

5

2. What is the probability of winning the jackpot? The total number of possible combinations is given by ⁵²C₆ = 52! / (6! * 46!) = 20,358,520 The probability of winning the jackpot with one ticket is therefore 1 / 20,358,520 ≈ 0.000000049 (extremely low!). This shows that even though people win the Lotto, the probability is incredibly small. Understanding this helps people make informed decisions about whether to spend their money on lottery tickets.

Example 3: Road Safety Suppose statistics show that 10% of drivers in a certain area speed and that speeding drivers are 4 times more likely to be involved in an accident than non-speeding drivers. What's the overall probability of being involved in an accident? Let S = Speeding Driver, NS = Non-Speeding Driver, A = Accident P(S) = 0.1 P(NS) = 0.9 Let P(A|NS) = x Then P(A|S) = 4x The overall probability of an accident is P(A) = P(A|S) P(S) + P(A|NS) P(NS) P(A) = (4x 0.1) + (x 0.9) = 0.4x + 0.9x = 1.3x To solve for x, we'd need additional information, such as the overall accident rate for the general driving population.