Welcome to Probability for Further Mathematics (AS 2 Section C)
Probability is the mathematical language of uncertainty. Whether predicting weather patterns, assessing medical diagnoses, or modelling financial markets, probability gives us the tools to quantify chance and make smart decisions. Don't worry if probability felt tricky in GCSE or standard A Level Maths; here we will build up the ideas step by step, using clear logic, visual diagrams, and practical examples so you feel totally confident for your CCEA AS 2 exam.
What you will learn in this chapter:
• The core rules of probability (Addition and Multiplication rules)
• How to work with mutually exclusive and independent events
• Mastering conditional probability using formulas, Venn diagrams, and tree diagrams
• The powerful Law of Total Probability and Bayes' Theorem
1. The Basics: Sample Space, Events, and Notation
Let's quickly refresh the foundational building blocks:
• Sample Space (\(S\) or \(\Omega\)): The set of all possible outcomes of an experiment. The total probability of the entire sample space is always \(P(S) = 1\).
• Event (\(A\)): A subset of the sample space (a specific collection of outcomes we care about).
• Probability bounds: For any event \(A\), \(0 \le P(A) \le 1\).
• Complement (\(A'\) or \(A^c\)): The event that \(A\) does not occur. Since something must either happen or not happen, \(P(A') = 1 - P(A)\).
Set Operations: "AND" vs "OR"
• Intersection (\(A \cap B\)): Event \(A\) AND event \(B\) both occur at the same time.
• Union (\(A \cup B\)): Event \(A\) OR event \(B\) occurs (or both occur).
Memory Trick: \(\cup\) looks like a cup that holds everything (Union = "all together" / OR), while \(\cap\) looks like an arch or an "n" for "and" (Intersection).
Key Takeaway
Every probability lies between \(0\) and \(1\), and the probability of something not happening is always \(1\) minus the probability that it does happen: \(P(A') = 1 - P(A)\).
---2. Combining Events: The Addition Rule
When we want to find the probability that either event \(A\) or event \(B\) happens, we look at the union \(P(A \cup B)\).
The General Addition Rule
If you add \(P(A)\) and \(P(B)\), you count the overlap (where both happen together) twice! To fix this, we subtract the overlap:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
Mutually Exclusive Events
Two events are mutually exclusive if they cannot happen at the same time. Think of rolling a standard dice: you cannot roll a \(2\) and a \(5\) on a single roll.
For mutually exclusive events:
• \(P(A \cap B) = 0\)
• The addition rule simplifies to: \(P(A \cup B) = P(A) + P(B)\)
Real-World Analogy: Turning left and turning right at a junction are mutually exclusive actions; you cannot do both simultaneously.
Quick Example: Venn Diagram Calculation
Suppose \(P(A) = 0.6\), \(P(B) = 0.5\), and \(P(A \cap B) = 0.3\).
Using the addition rule:
\(P(A \cup B) = 0.6 + 0.5 - 0.3 = 0.8\)
The probability of neither event occurring is \(P((A \cup B)') = 1 - 0.8 = 0.2\).
Key Takeaway
Always check whether events can occur together. If they can overlap, you must subtract \(P(A \cap B)\) when calculating \(P(A \cup B)\).
---3. Conditional Probability and Independence
Conditional probability is the probability of an event occurring given that another event is already known to have occurred. We write this as \(P(A|B)\), read as "the probability of \(A\) given \(B\)".
The Conditional Probability Formula
When we know \(B\) has happened, our entire "universe" shrinks from the whole sample space down to just \(B\). Within this new universe, the only way \(A\) can happen is if both \(A\) and \(B\) happen:
\(P(A|B) = \frac{P(A \cap B)}{P(B)}\), provided \(P(B) > 0\)
Rearranging this formula gives the Multiplication Rule:
\(P(A \cap B) = P(A|B) \times P(B) = P(B|A) \times P(A)\)
Independent Events
Two events \(A\) and \(B\) are independent if the occurrence of one has no effect whatsoever on the probability of the other.
Mathematically, \(A\) and \(B\) are independent if and only if:
\(P(A|B) = P(A)\) (knowing \(B\) occurred changes nothing about \(A\))
which leads directly to the multiplication test for independence:
\(P(A \cap B) = P(A) \times P(B)\)
Crucial Warning: Independent vs Mutually Exclusive
Students often mix these two concepts up! Let's clear this up once and for all:
• Mutually Exclusive: Cannot happen together (\(P(A \cap B) = 0\)). If one happens, the other cannot happen.
• Independent: Can happen together, but one does not influence the other (\(P(A \cap B) = P(A) \times P(B)\)).
• Did you know? If two non-impossible events are mutually exclusive, they cannot be independent, because knowing one happened tells you for certain that the other did not!
Key Takeaway
To prove two events are independent in an exam, always calculate \(P(A) \times P(B)\) and show whether or not it equals \(P(A \cap B)\).
---4. Using Two-Way Tables and Tree Diagrams
Visual representations make conditional probability much easier to solve without getting lost in algebraic formulas.
Two-Way Contingency Tables
A two-way table organizes counts or probabilities for two categorical variables.
Example: A group of \(100\) students is surveyed about studying Further Maths (\(M\)) and Physics (\(F\)).
• \(40\) study both Further Maths and Physics.
• \(10\) study Further Maths but not Physics.
• \(30\) study Physics but not Further Maths.
• \(20\) study neither.
From these counts:
• Total studying Further Maths: \(P(M) = \frac{40 + 10}{100} = \frac{50}{100} = 0.5\)
• Total studying Physics: \(P(F) = \frac{40 + 30}{100} = \frac{70}{100} = 0.7\)
• Probability a randomly chosen Physics student also studies Further Maths: \(P(M|F) = \frac{P(M \cap F)}{P(F)} = \frac{40/100}{70/100} = \frac{40}{70} = \frac{4}{7}\).
Tree Diagrams
Tree diagrams are ideal for sequential stages or experiments with multiple steps.
• First set of branches: Show marginal probabilities \(P(B)\) and \(P(B')\).
• Second set of branches: Show conditional probabilities like \(P(A|B)\) and \(P(A|B')\).
• Multiply along branches: Gives the intersection probabilities, e.g., \(P(B \cap A) = P(B) \times P(A|B)\).
• Add across paths: Gives the total probability of an outcome arriving via different routes.
Key Takeaway
When given counts or survey data, sketch a two-way table. When given a chain of events or conditional stages, draw a tree diagram and multiply along the branches.
---5. The Law of Total Probability & Bayes' Theorem
In Further Mathematics, you will solve problems where an outcome can occur through several distinct scenarios. This is where the Law of Total Probability and Bayes' Theorem become your most valuable tools.
Partitions of a Sample Space
A set of events \(B_1, B_2, \dots, B_k\) forms a partition of the sample space \(S\) if:
1. They are pairwise mutually exclusive (no overlap: \(B_i \cap B_j = \emptyset\) for \(i \neq j\)).
2. They are exhaustive (they cover everything: \(B_1 \cup B_2 \cup \dots \cup B_k = S\)).
The Law of Total Probability
If \(B_1, B_2, \dots, B_k\) partition the sample space, then any event \(A\) can be split into pieces based on which \(B_i\) it occurs with:
\(P(A) = P(A \cap B_1) + P(A \cap B_2) + \dots + P(A \cap B_k)\)
Applying the multiplication rule to each term gives:
\(P(A) = \sum_{i=1}^{k} P(A|B_i)P(B_i)\)
For two partitions (\(B\) and \(B'\)), this is simply:
\(P(A) = P(A|B)P(B) + P(A|B')P(B')\)
Bayes' Theorem
Bayes' Theorem allows us to "reverse" conditional probabilities. If we know the forward conditional probability \(P(A|B_i)\), Bayes' Theorem lets us find the backward probability \(P(B_i|A)\):
\(P(B_i|A) = \frac{P(A \cap B_i)}{P(A)} = \frac{P(A|B_i)P(B_i)}{\sum_{j=1}^{k} P(A|B_j)P(B_j)}\)
For the standard two-event case:
\(P(B|A) = \frac{P(A|B)P(B)}{P(A|B)P(B) + P(A|B')P(B')}\)
Step-by-Step Worked Example: Diagnostic Testing
A rare disease affects \(1\%\) of a population. A clinic develops a test for the disease:
• If a person has the disease (\(D\)), the test is positive (\(T^+\)) with probability \(0.98\).
• If a person does not have the disease (\(D'\)), the test is falsely positive with probability \(0.05\).
A randomly selected person tests positive. What is the probability they actually have the disease?
Step 1: Identify the given probabilities
• Prior probability of disease: \(P(D) = 0.01 \implies P(D') = 0.99\)
• True positive rate: \(P(T^+|D) = 0.98\)
• False positive rate: \(P(T^+|D') = 0.05\)
Step 2: Find total probability of a positive test \(P(T^+)\) using Total Probability
\(P(T^+) = P(T^+|D)P(D) + P(T^+|D')P(D')\)
\(P(T^+) = (0.98 \times 0.01) + (0.05 \times 0.99)\)
\(P(T^+) = 0.0098 + 0.0495 = 0.0593\)
Step 3: Apply Bayes' Theorem to find \(P(D|T^+)\)
\(P(D|T^+) = \frac{P(T^+|D)P(D)}{P(T^+)} = \frac{0.0098}{0.0593} \approx 0.1653 \text{ (or } 16.5\%)\)
Surprising result! Even with a positive test, there is only about a \(16.5\%\) chance the person has the disease. This is because the disease is so rare in the general population that false positives outnumber true positives!
Key Takeaway
Bayes' Theorem updates our prior belief \(P(B)\) in light of new evidence \(A\) to produce the posterior probability \(P(B|A)\). Always find the total probability of the evidence \(P(A)\) in the denominator first.
---6. Common Pitfalls and Exam Tips
• Mixing up the condition: Be careful not to confuse \(P(A|B)\) with \(P(B|A)\). The condition (the event that has already occurred) always goes after the vertical line and in the denominator.
• Assuming independence: Never assume \(P(A \cap B) = P(A) \times P(B)\) unless the question explicitly states the events are independent or you are asked to test for it.
• Probabilities exceeding 1: If your calculation yields a probability greater than \(1\) or less than \(0\), stop and review your steps. An overlap was likely double-counted.
• Clear tree diagram labelling: When drawing tree diagrams, label every branch clearly with both the event name and the conditional probability. Ensure the probabilities on any set of branching paths add up to \(1\).
Summary Checklist
Before sitting your exam, make sure you can:
1. State and apply the general addition formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
2. State and apply the conditional probability formula: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
3. Test for independence using \(P(A \cap B) = P(A)P(B)\) or \(P(A|B) = P(A)\)
4. Use the Law of Total Probability to find overall probabilities across multiple scenarios
5. Apply Bayes' Theorem accurately to reverse conditional probabilities