Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Define and calculate conditional probabilities.
• Determine if two events are independent.
• Solve real-world problems involving these concepts.
• Use tree diagrams and contingency tables for analysis.
• Differentiate between independent and dependent events.

Lesson summary

Probability is not just about tossing coins or rolling dice; it’s a fundamental tool for understanding and predicting the likelihood of events around us. In South Africa, probability plays a crucial role in diverse areas, from predicting weather patterns for agriculture and disaster preparedness to understanding the risks involved in investments and insurance. Understanding probability empowers us to make informed decisions in our daily lives. This week, we delve into conditional probability, which helps us analyze events that are influenced by prior occurrences, and independent events, where one event has no effect on the other.

Lesson notes

2.1 Conditional Probability: Conditional probability is the probability of an event A occurring, given that another event B has already occurred. This is written as P(A|B), which is read as "the probability of A given B." The formula for 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 given event B. P(A ∩ B) is the probability of both events A and B occurring. This is the intersection of A and B. P(B) is the probability of event B occurring. Why does this formula work? We're effectively shrinking our sample space. We know B has already happened.

Therefore, we are only interested in the portion of event A that overlaps with event B (A ∩ B). We then divide this probability by the probability of event B itself to get the conditional probability.

Example 1: Voter Registration in Gauteng Suppose a survey in Gauteng showed that 60% of adults are registered to vote, and 85% of those registered to vote are employed. What is the probability that a randomly selected adult in Gauteng is both registered to vote and employed? Let A be the event "registered to vote" and B be the event "employed." We are given P(A) = 0.60 and P(B|A) = 0.

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5. We need to find P(A ∩ B). Using the conditional probability formula, rearranged: P(A ∩ B) = P(B|A) * P(A) P(A ∩ B) = 0.85 * 0.60 P(A ∩ B) = 0.51 Therefore, the probability that a randomly selected adult in Gauteng is both registered to vote and employed is 51%.

Example 2: Load Shedding and Business Closure A small business owner in Cape Town estimates that the probability of load shedding on any given day is 0.

3. If load shedding occurs, the probability that the business will have to close for the day is 0.

8. What is the probability that the business will close on any given day? Let A be the event "business closes" and B be the event "load shedding occurs." We are given P(B) = 0.3 and P(A|B) = 0.

8. We need to find P(A ∩ B). P(A ∩ B) = P(A|B) * P(B) P(A ∩ B) = 0.8 * 0.3 P(A ∩ B) = 0.24 Therefore, the probability that the business will close on any given day is 24%. 2.2 Independent Events: Two events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, A and B are independent if: P(A|B) = P(A) or P(B|A) = P(B) An easier (and more common) test for independence is: P(A ∩ B) = P(A) * P(B) Why does this work? If A and B are independent, knowing that B happened doesn't change the probability of

A. So the probability of A given B is simply the probability of

A. Example 3: Coin Toss and Dice Roll Consider tossing a fair coin and rolling a fair six-sided die. The coin toss results (Heads or Tails) are independent of the die roll results (1, 2, 3, 4, 5, or 6). Let A be the event "getting Heads on the coin toss" and B be the event "rolling a 3 on the die." P(A) = 1/2 P(B) = 1/6 P(A ∩ B) = P(A) P(B) = (1/2) (1/6) = 1/12 This confirms that the events are independent.

Example 4: Drawing Cards with and without Replacement Consider drawing two cards from a standard deck of 52 cards.

With Replacement: If you draw the first card, replace it, and then draw the second card, the two draws are independent events because the outcome of the first draw does not affect the probabilities of the second draw.

Without Replacement: If you draw the first card and do not replace it, the two draws are dependent events. The outcome of the first draw changes the composition of the deck, thereby altering the probabilities of the second draw. Let A be the event "drawing an Ace on the first draw" and B be the event "drawing an Ace on the second draw." With Replacement: P(A) = 4/52 = 1/

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3. P(B) = 4/52 = 1/

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3. P(A ∩ B) = (1/13) (1/13) = 1/

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9. Events are independent.

Without Replacement: P(A) = 4/52 = 1/

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3. P(B|A) = 3/51 = 1/

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7. The probability of drawing a second Ace, given that the first card was an Ace, is 3/

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1. Events are dependent. P(A ∩ B) = (4/52)(3/51)=1/221 2.3 Tree Diagrams and Contingency Tables: Tree Diagrams: These are useful for visualizing and calculating probabilities in sequential events, especially when dealing with conditional probabilities. Each branch represents a possible outcome, and probabilities are written along the branches. To find the probability of a sequence of events, multiply the probabilities along the corresponding branches.

Contingency Tables: These are tables that display the frequency distribution of two or more categorical variables. They are particularly helpful in calculating probabilities, especially conditional probabilities, because they organize the data in a way that makes it easy to see the relationships between different events. Guided Practice (With Solutions)

Question 1: In a class of 40 students, 25 study Mathematics and 18 study Physics. 10 students study both Mathematics and Physics. What is the probability that a randomly selected student studies Physics given that they study Mathematics?