Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Probability cont
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Probability cont
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 3
Theme: Statistics
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This topic builds upon the foundational concepts of probability introduced in previous weeks, focusing on the calculation of probabilities for compound events, specifically independent and dependent events, and the effective use of tree diagrams. Understanding these concepts is crucial for students as they will encounter situations in real life where outcomes of multiple events need to be analysed to make informed decisions. In Nigeria, this knowledge is applicable in various sectors such as risk assessment in business and insurance, predicting outcomes in sports, evaluating public health interventions, and understanding election projections.
This section extends the understanding of probability to scenarios involving multiple events. 2.
1. Review of Basic Probability Probability (P(A)): The likelihood of an event A occurring, expressed as a fraction, decimal, or percentage. $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Sample Space (S): The set of all possible outcomes of an experiment.
Event: A subset of the sample space.
Mutually Exclusive Events: Events that cannot occur at the same time. For such events, $P(A \text{ or } B) = P(A) + P(B)$.
Complementary Events: Two events where one event occurring means the other cannot. $P(A) + P(A') = 1$, where A' is the complement of A. 2.
2. Independent Events Definition: Two events, A and B, are independent if the occurrence of one does not affect the probability of the other occurring.
Example: Tossing a coin twice. The outcome of the first toss (Heads or Tails) does not influence the outcome of the second toss. Multiplication Rule for Independent Events: If A and B are independent events, the probability that both A and B occur is given by: $P(A \text{ and } B) = P(A) \times P(B)$ This rule can be extended to more than two independent events: $P(A \text{ and } B \text{ and } C) = P(A) \times P(B) \times P(C)$ Worked Example 2.2.1 (Nigerian Context): A Nigerian farmer plants yam and maize in separate fields. The probability that the yam harvest is good is 0.8, and the probability that the maize harvest is good is 0.
7. Assuming the success of one crop does not affect the other, calculate the probability that: a) Both harvests are good. b) Neither harvest is good.
Solution: Let Y be the event of a good yam harvest, and M be the event of a good maize harvest.
Given: $P(Y) = 0.8$, $P(M) = 0.7$.
Since the events are independent: a)
Probability that both harvests are good: $P(Y \text{ and } M) = P(Y) \times P(M)$ $P(Y \text{ and } M) = 0.8 \times 0.7 = 0.56$ b)
Probability that neither harvest is good: First, find the probability that yam harvest is not good: $P(Y') = 1 - P(Y) = 1 - 0.8 = 0.2$. Next, find the probability that maize harvest is not good: $P(M') = 1 - P(M) = 1 - 0.7 = 0.3$.
Probability that neither harvest is good: $P(Y' \text{ and } M') = P(Y') \times P(M')$ $P(Y' \text{ and } M') = 0.2 \times 0.3 = 0.06$ 2.
3. Dependent Events Definition: Two events, A and B, are dependent if the occurrence of one event affects the probability of the other occurring.
Example: Drawing two cards from a deck without replacement. The probability of drawing a certain card on the second draw depends on what card was drawn first. Conditional Probability (P(B|A)): The probability of event B occurring given that event A has already occurred.
Multiplication Rule for Dependent Events: If A and B are dependent events, the probability that both A and B occur is given by: $P(A \text{ and } B) = P(A) \times P(B|A)$ This rule can also be extended: $P(A \text{ and } B \text{ and } C) = P(A) \times P(B|A) \times P(C|A \text{ and } B)$ Worked Example 2.3.1 (Nigerian Context): A basket contains 5 ripe mangoes and 3 unripe mangoes. A student from the class picks two mangoes at random, one after the other, without replacement.
Calculate the probability that: a) Both mangoes picked are ripe. b) The first mango is ripe and the second is unripe.
Solution: Total mangoes = 5 (ripe) + 3 (unripe) = 8 mangoes. a) Probability that both mangoes picked are ripe: Let R1 be the event that the first mango is ripe. $P(R1) = \frac{\text{Number of ripe mangoes}}{\text{Total mangoes}} = \frac{5}{8}$ After picking one ripe mango, there are now 4 ripe mangoes left and 7 total mangoes. Let R2 be the event that the second mango is ripe, given the first was ripe. $P(R2|R1) = \frac{4}{7}$ Probability of both being ripe: $P(R1 \text{ and } R2) = P(R1) \times P(R2|R1)$ $P(R1 \text{ and } Total mangoes = 5 (ripe) + 3 (unripe) = 8 mangoes. a) Probability that both mangoes picked are ripe: Let R1 be the event that the first mango is ripe. $P(R1) = \frac{\text{Number of ripe mangoes}}{\text{Total mangoes}} = \frac{5}{8}$ After picking one ripe mango, there are now 4 ripe mangoes left and 7 total mangoes. Let R2 be the event that the second mango is ripe, given the first was ripe. $P(R2|R1) = \frac{4}{7}$ Probability of both being ripe: $P(R1 \text{ and } R2) = P(R1) \times P(R2|R1)$ $P(R1 \text{ and } R2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ b) Probability that the first mango is ripe and the second is unripe: Let R1 be the event that the first mango is ripe. $P(R1) = \frac{5}{8}$ After picking one ripe mango, there are 4 ripe mangoes and 3 unripe mangoes left (total 7 mangoes). Let U2 be the event that the second mango is unripe, given the first was ripe. $P(U2|R1) = \frac{3}{7}$ Probability of first ripe and second unripe: $P(R1 \text{ and } U2) = P(R1) \times P(U2|R1)$ $P(R1 \text{ and } U2) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$ 2.
4. Tree Diagrams Purpose: Tree diagrams are visual tools used to represent the sequence of events and their probabilities. They are particularly useful for problems involving two or more consecutive events, especially dependent events.
Construction:
1. Start with a point representing the beginning of the experiment.
2. Draw branches for each possible outcome of the first event. Write the probability of each outcome on its branch.
3. From the end of each first-event branch, draw further branches for the possible outcomes of the second event, and write their conditional probabilities on these branches.
4. Continue this process for subsequent events.
5. The ends of the final branches represent all possible combined outcomes.
Interpretation: To find the probability of a specific sequence of events, multiply the probabilities along the branches leading to that outcome. To find the probability of an event that can occur in several ways, add the probabilities of all the relevant sequences (final outcomes). Worked Example 2.4.1 (Nigerian Context): A local bakery makes two types of bread: white and brown. 60% of their bread is white, and 40% is brown. 80% of the white bread is sold by noon, while 70% of the brown bread is sold by noon. Construct a tree diagram and use it to find the probability that a randomly chosen loaf of bread is: a) White bread and sold by noon. b) Not sold by noon.
Solution: Let W be White bread, B be Brown bread. Let S be Sold by noon, S' be Not sold by noon.
First event (Type of bread): $P(W) = 0.60$ $P(B) = 0.40$ Second event (Sold by noon), conditional on bread type: $P(S|W) = 0.80$ (80% of white bread is sold) $P(S'|W) = 1 - 0.80 = 0.20$ (20% of white bread is not sold) $P(S|B) = 0.70$ (70% of brown bread is sold) $P(S'|B) = 1 - 0.70 = 0.30$ (30% of brown bread is not sold)
Tree Diagram: ``` Start | | +----|----+ | | P(W)=0.6 P(B)=0.4 | | W B | | +--|--+ +--|--+ | | | | P(S|W) P(S'|W) P(S|B) P(S'|B) =0.8 =0.2 =0.7 =0.3 | | | | S S' S S' (W,S) (W,S') (B,S) (B,S')
Outcomes and Probabilities: P(W and S) = P(W) P(S|W) = 0.6 0.8 = 0.48 P(W and S') = P(W) P(S'|W) = 0.6 0.2 = 0.12 P(B and S) = P(B) P(S|B) = 0.4 0.7 = 0.28 P(B and S') = P(B) P(S'|B) = 0.4 * 0.3 = 0.12 (Sum of probabilities = 0.48 + 0.12 + 0.28 + 0.12 = 1.00) ``` a) Probability that a randomly chosen loaf is white bread and sold by noon: From the tree diagram, $P(W \text{ and } S) = 0.48$. b) Probability that a randomly chosen loaf is not sold by noon: This can happen in two ways: (White bread AND not sold) OR (Brown bread AND not sold). $P(\text{not sold by noon}) = P(W \text{ 0.7 = 0.28 P(B and S') = P(B) P(S'|B) = 0.4 * 0.3 = 0.12 (Sum of probabilities = 0.48 + 0.12 + 0.28 + 0.12 = 1.00) ``` a) Probability that a randomly chosen loaf is white bread and sold by noon: From the tree diagram, $P(W \text{ and } S) = 0.48$. b) Probability that a randomly chosen loaf is not sold by noon: This can happen in two ways: (White bread AND not sold) OR (Brown bread AND not sold). $P(\text{not sold by noon}) = P(W \text{ and } S') + P(B \text{ and } S')$ $P(\text{not sold by noon}) = 0.12 + 0.12 = 0.24$ --- 3.
1. Introduction (10 minutes)
Teacher Activity: Begins by reviewing basic probability concepts through a quick question-and-answer session. Asks students to define probability, sample space, and give an example of mutually exclusive events. Explains that this week's lesson builds on these foundations to explore events that happen sequentially or simultaneously.
Student Activity: Students respond to questions, recalling prior knowledge. 3.
2. Concept Development: Independent Events (15 minutes)
Teacher Activity: Introduces the concept of independent events with relatable examples (e.g., tossing two coins, rolling two dice, the outcome of two separate football matches in the Nigerian Professional Football League). Defines the multiplication rule for independent events: $P(A \text{ and } B) = P(A) \times P(B)$. Works through Example 2.2.1 on the board, explaining each step clearly.
Student Activity: Students listen, take notes, ask clarifying questions, and participate by suggesting examples of independent events. 3.
3. Concept Development: Dependent Events (20 minutes)
Teacher Activity: Introduces the concept of dependent events, contrasting it with independent events. Uses examples like drawing cards without replacement, selecting students from a class for a specific task. Explains conditional probability $P(B|A)$ as the probability of B given
A. Introduces the multiplication rule for dependent events: $P(A \text{ and } B) = P(A) \times P(B|A)$. Works through Example 2.3.1 on the board, emphasizing how the sample space changes after the first event.
Student Activity: Students engage in active listening, observe the worked example, and attempt to explain the difference between independent and dependent events in their own words. 3.
4. Tree Diagrams (25 minutes)
Teacher Activity: Introduces tree diagrams as a visual method to solve probability problems involving sequences of events. Demonstrates step-by-step how to construct a tree diagram using Example 2.4.1 (the bakery example) on the board. Shows how to label branches with probabilities and how to calculate the probability of combined outcomes by multiplying along the branches. Explains how to sum probabilities of different paths to find the probability of a more complex event (e.g., "not sold by noon").
Student Activity: Students pay close attention to the construction process. They draw along with the teacher, ensuring they understand the structure and calculations involved. 3.
5. Group Work & Problem Solving (15 minutes)
Teacher Activity: Divides the class into small groups (3-4 students). Distributes a few practice problems that require identifying event types (independent/dependent) and/or constructing tree diagrams. Circulates to provide support, guidance, and check for understanding.
Student Activity: Students work collaboratively in their groups to solve the assigned problems, discussing strategies and calculations. They prepare to present their solutions. 3.
6. Class Discussion and Wrap-up (5 minutes)
Teacher Activity: Selects one or two groups to share their solutions and explanations. Addresses any misconceptions or common errors observed during group work. Summarizes the key learning points of the lesson.
Student Activity: Students present their findings. They listen to other groups' solutions and participate in a brief recap of the lesson. ---
Agriculture and Weather Forecasting (Nigerian context): Farmers in Nigeria use probability to make decisions. The probability of good rainfall can be combined with the probability of pest infestation (often independent) to assess the overall probability of a successful harvest for various crops like maize, rice, or yam. For instance, a farmer might calculate the probability of having both sufficient rainfall and minimal pest damage, which are independent events that impact yield. Public Health and Disease Management (Nigerian context): Public health officials and epidemiologists in Nigeria use probability to predict the spread of diseases like malaria, cholera, or Lassa fever. They might consider the probability of an individual being exposed to a vector (e.g., mosquito) and the conditional probability of developing the disease given exposure. Tree diagrams can model the sequence of events (exposure -> infection -> recovery/death) to estimate population-level outcomes or the effectiveness of interventions like vaccination campaigns.
Sports and Betting (Nigerian context): The Nigerian sports industry, especially football, extensively uses probability. Sport analysts and bettors calculate the probability of a specific Nigerian football club winning a match, or the probability of multiple teams winning their respective matches (independent events). The probability of a particular player scoring depends on various factors, and combined probabilities can be used to predict game outcomes or analyze player performance. Business and Quality Control (Nigerian context): Manufacturing companies in Nigeria use probability for quality control. For example, a textile factory in Kano producing Ankara fabrics might sample fabrics for defects. The probability of selecting two defective fabrics consecutively (without replacement) indicates the overall quality of a batch and informs decisions on process improvement. Probability is also used for insurance premium calculations, assessing the likelihood of claims based on various independent and dependent risk factors. ---