Probability: predicting outcomes and risk – Week 6 focus
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Subject: Mathematical Literacy
Class: Grade 11
Term: Term 4
Week: 6
Theme: General lesson support
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This week, we delve into the exciting world of probability, focusing on how we can use probability to predict outcomes and understand risks. Understanding probability is crucial in making informed decisions in our daily lives. From assessing the likelihood of winning a lottery to understanding the risks associated with driving in different weather conditions or investing in different schemes, probability helps us navigate uncertainty. In South Africa, where factors like crime, health risks, and economic volatility are significant, a solid grasp of probability empowers us to make more reasoned choices.
2.1 Basic Probability Review: Probability is the measure of the likelihood that an event will occur. It's expressed as a number between 0 and 1, where 0 indicates impossibility, and 1 indicates certainty.
The basic formula for probability is: Probability of Event A (P(A)) = (Number of favorable outcomes for A) / (Total number of possible outcomes)
Example 1: Tossing a fair coin. What is the probability of getting heads?
Solution: There's 1 favorable outcome (heads) and 2 possible outcomes (heads or tails). P(Heads) = 1/2 = 0.5 or 50% Example 2: Rolling a fair six-sided die. What is the probability of rolling a 4?
Solution: There's 1 favorable outcome (rolling a 4) and 6 possible outcomes (1, 2, 3, 4, 5, 6). P(Rolling a 4) = 1/6 ≈ 0.167 or 16.7% 2.2 Types of Events: Independent Events: The outcome of one event does not affect the outcome of another event.
For independent events A and B: P(A and B) = P(A)
P(B)
Example 3: Flipping a coin twice. The first flip doesn't affect the second. What's the probability of getting heads twice in a row?
Solution: P(Heads on first flip) = 1/
2. P(Heads on second flip) = 1/
2. P(Heads and Heads) = (1/2) * (1/2) = 1/4 = 0.25 or 25% Dependent Events: The outcome of one event does affect the outcome of another event.
For dependent events A and B: P(A and B) = P(A) P(B|A), where P(B|A) is the probability of B given that A has already occurred.
Example 4: Drawing two cards from a deck without replacement. What's the probability of drawing two Aces in a row?
Solution: P(Ace on first draw) = 4/52 (there are 4 Aces in a deck of 52 cards). P(Ace on second draw, given an Ace was drawn first) = 3/51 (now there are only 3 Aces left and 51 total cards). P(Ace and Ace) = (4/52) * (3/51) = 12/2652 = 1/221 ≈ 0.0045 or 0.45% Mutually Exclusive Events: Two events that cannot happen at the same time. If A and B are mutually exclusive, P(A and B) =
0. For mutually exclusive events A and B: P(A or B) = P(A) + P(B)
Example 5: Rolling a die. You can't roll a 3 and a 5 at the same time. What is the probability of rolling a 3 or a 5?
Solution: P(Rolling a 3) = 1/
6. P(Rolling a 5) = 1/
6. P(Rolling a 3 or 5) = (1/6) + (1/6) = 2/6 = 1/3 ≈ 0.333 or 33.3% Non-Mutually Exclusive Events: Two events that can happen at the same time.
For non-mutually exclusive events A and B: P(A or B) = P(A) + P(B) - P(A and B)
Example 6: Drawing a card from a deck. What's the probability of drawing a heart or a king?
Solution: P(Heart) = 13/
5
2. P(King) = 4/
5
2. P(Heart and King) = 1/52 (the King of Hearts). P(Heart or King) = (13/52) + (4/52) - (1/52) = 16/52 = 4/13 ≈ 0.308 or 30.8% 2.3 Probability Tools: Tree Diagrams & Venn Diagrams: Tree Diagrams: Useful for visualizing sequential events and their probabilities, especially when dealing with dependent events. Each branch represents a possible outcome, and the probabilities are written along the branches. To find the probability of a sequence of events, you multiply the probabilities along the corresponding branches.
Venn Diagrams: Useful for visualizing the relationships between sets of events, especially when dealing with mutually exclusive or non-mutually exclusive events. The overlapping region represents the intersection of the events (the outcomes that are common to both events). 2.4 Expected Value: The expected value (EV) is the average outcome you would expect if you repeated an experiment many times. It's calculated by multiplying each possible outcome by its probability and then summing those products. EV = (Outcome 1 Probability 1) + (Outcome 2 Probability 2) + ... + (Outcome n * Probability n)
Example 7: A raffle has 1 prize of R1000 and 10 prizes of R50. 100 tickets are sold at R20 each. What is the expected value of buying one ticket?
Solution: Probability of winning R1000: 1/100 Probability of winning R50: 10/100 = 1/10 Probability of winning nothing: 89/100 However, remember that you spent R20 on the ticket, so we must deduct that. EV = (R1000 1/100) + (R50 1/10) + (R0 * 89/100) - R20 EV = R10 + R5 + R0 - R20 EV = -R5 This means that, on average, you are expected to lose R5 for each ticket you buy. This is why raffles are generally unfavorable for the ticket buyer. 2.5 Risk Assessment: Probability is fundamental to assessing risk. Higher probability of a negative outcome translates to higher risk. We use probability to quantify and understand the likelihood of adverse events, enabling us to make informed decisions to mitigate those risks.
Example 8: Home insurance. The insurance company calculates the probability of various events (fire, theft, flood, etc.) occurring at your property. Based on these probabilities and the potential costs associated with each event, they determine the premium you pay. Higher-risk areas (e.g., flood plains, high-crime areas) will have higher premiums because the probability of a claim is higher. Guided Practice (With Solutions)
Question 1: A bag contains 5 red balls and 3 blue balls.