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IB Mathematics AA Venn and tree diagrams, counting principles Study Notes

IB Mathematics AA Venn and tree diagrams, counting principles Study Notes

IB Mathematics AA Venn and tree diagrams, counting principles Study Notes

IB Mathematics AA Venn and tree diagrams, counting principles Notes Offer a clear explanation of Venn and tree diagrams, counting principles, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Venn and tree diagrams, counting principles.

Venn and Tree Diagrams, Counting Principles

Introduction

Probability problems often involve multiple events and their relationships. Venn diagrams, tree diagrams, sample space diagrams, and outcome tables are visual tools that help represent and solve these problems. Understanding concepts like combined events, conditional probability, and independent events is crucial in this context.

Key Concepts

1. Venn Diagrams
  • Definition: A graphical representation of sets and their relationships, such as intersection, union, and complement.
  • Formula for Combined Events:

    \( P(A \cup B) = P(A) + P(B) – P(A \cap B) \)

  • Special Case – Mutually Exclusive Events:

    \( P(A \cap B) = 0 \)

    Events \( A \) and \( B \) cannot occur simultaneously.
2. Tree Diagrams
  • Definition: A branching diagram representing all possible outcomes of a sequence of events.
  • Use: Useful for probabilities with and without replacement.

3. Conditional Probability
  • Definition: The probability of event \( A \) occurring, given that event \( B \) has occurred.

    \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)

  • Alternate Form:

    \( P(A \cap B) = P(B)P(A|B) \)

4. Independent Events
  • Definition: Events \( A \) and \( B \) are independent if the occurrence of one does not affect the probability of the other.
  • Formula:

    \( P(A \cap B) = P(A)P(B) \)

5. Counting Principles
  • Sample Space: Can be represented using tables, lists, or diagrams.
  • Fundamental Counting Principle: If one event can occur in \( m \) ways and a second event in \( n \) ways, then the total number of outcomes is \( m \cdot n \).

Solved Examples

Example 1: Combined Events Using Venn Diagram

Problem: In a class of 40 students, 25 study mathematics, 20 study physics, and 10 study both. Find the probability that a randomly chosen student studies either mathematics or physics.

Solution:

  • Total students: \( n(U) = 40 \)

  • Number of students studying mathematics (\( n(A) \)): 25
  • Number of students studying physics (\( n(B) \)): 20
  • Number studying both (\( n(A \cap B) \)): 10
  • Using the formula:

    \( P(A \cup B) = P(A) + P(B) – P(A \cap B) \)

    \( P(A \cup B) = \frac{25}{40} + \frac{20}{40} – \frac{10}{40} = \frac{35}{40} = 0.875 \)

Answer: The probability is 0.875 or 87.5%.

Example 2: Conditional Probability

Problem: The probability of passing mathematics is 0.8, and the probability of passing both mathematics and physics is 0.5. What is the probability of passing physics given that a student has passed mathematics?

Solution:

  • Given:
    • \( P(A) = 0.8 \)
    • \( P(A \cap B) = 0.5 \)
  • Using the formula:

    \( P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.5}{0.8} = 0.625 \)

Answer: The probability is 0.625 or 62.5%.

Applications and Connections

  • Real-World Applications:
    • Gambling: Understanding odds and probabilities.
    • Insurance: Risk assessment and premium calculation.
  • Ethical Considerations:
    • Gambling ethics: Should mathematics be used to optimize gambling strategies?
    • Responsibility of mathematicians in applying their knowledge.

IB Mathematics AA SL Venn and tree diagrams, counting principles Exam Style Worked Out Questions

Question

At Nusaybah’s Breakfast Diner, three types of omelette are available to order: chicken, vegetarian and steak. Each omelette is served with either a portion of fries or hash browns. It is known that 20 % of customers choose a chicken omelette, 70 % choose a vegetarian omelette and 10 % choose a steak omelette.

It is also known that 65% of those ordering the chicken omelette, 70 % of those ordering the vegetarian omelette and 60 % of those ordering the steak omelette, order fries.

The following tree diagram represents the orders made by each customer.

           

  1. Complete the tree diagram by adding the respective probabilities to each branch. [2]

  2. Find the probability that a randomly selected customer orders fries. [2]

  3. Find the probability that a randomly selected customer orders fries, given that they do [Maximum mark: 7]

Answer/Explanation

Ans: 

(a)

(b)     (0.2 × 0.5)+(0.7 × 0.7)+(0.1× 0.6)= 0.68 \((=\frac{17}{25})\)

(c)

Question

Let A and B be events such that \({\text{P}}(A) = 0.6,{\text{ P}}(A \cup B) = 0.8{\text{ and P}}(A|B) = 0.6\) .

Find P(B) .

Answer/Explanation

Markscheme

EITHER

Using \({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\)     (M1)

\(0.6{\text{P}}(B) = {\text{P}}(A \cap B)\)     A1

Using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) to obtain \(0.8 = 0.6 + {\text{P}}(B) – {\text{P}}(A \cap B)\)     A1

Substituting \(0.6{\text{P}}(B) = {\text{P}}(A \cap B)\) into above equation     M1

OR

As \({\text{P}}(A|B) = {\text{P}}(A)\) then A and B are independent events     M1R1

Using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A) \times {\text{P}}(B)\)     A1

to obtain \(0.8 = 0.6 + {\text{P}}(B) – 0.6 \times {\text{P}}(B)\)     A1

THEN

\(0.8 = 0.6 + 0.4{\text{P}}(B)\)     A1

\({\text{P}}(B) = 0.5\)     A1     N1

[6 marks]

Examiners report

This question was generally well done, with a few candidates spotting an opportunity to use results for the independent events A and B.

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