Welcome to AS Probability!
Welcome to one of the most practical and exciting areas of AS 2: Applied Mathematics! Whether predicting the weather, assessing risks in finance, or testing new medicines, probability gives us the mathematical tools to make sense of chance and uncertainty. Don't worry if probability felt confusing in the past—we will build every concept step by step, using clear diagrams, straightforward rules, and real-world analogies.
Did you know? The mathematical study of probability began in the 17th century when French mathematicians Blaise Pascal and Pierre de Fermat exchanged letters trying to solve a gambling puzzle known as the "Problem of Points"!
1. Fundamental Ideas and Set Notation
Before diving into calculations, let's establish the common language and symbols used in CCEA AS Mathematics.
Key Definitions:
• Experiment: A repeatable process that produces an outcome (e.g., rolling a fair six-sided die).
• Sample Space (\(S\) or \(\xi\)): The set of all possible outcomes. For a standard die, \(S = \{1, 2, 3, 4, 5, 6\}\).
• Event: A collection of one or more outcomes (e.g., rolling an even number: \(E = \{2, 4, 6\}\)).
Probability Basics:
The probability of any event \(A\) always lies between \(0\) and \(1\) inclusive:
\(0 \le P(A) \le 1\)
• \(P(A) = 0\) means the event is impossible.
• \(P(A) = 1\) means the event is certain to happen.
• The sum of probabilities for all exhaustive, mutually exclusive outcomes is always \(1\): \(\sum P(X) = 1\).
Set Notation Symbols You Need to Know
• Complement (\(A'\)): The event that \(A\) does not happen.
Rule: \(P(A') = 1 - P(A)\)
• Intersection (\(A \cap B\)): The event that both \(A\) and \(B\) happen. Think of the overlapping region in a Venn diagram. (Memory trick: \(\cap\) looks like an 'n' for and).
• Union (\(A \cup B\)): The event that \(A\) or \(B\) or both happen. (Memory trick: \(\cup\) looks like a 'u' for union or 'o' for or).
Key Takeaway: Always check that your probabilities never exceed \(1\) or drop below \(0\). The complement rule \(P(A') = 1 - P(A)\) is one of your handiest tools!
2. The Addition Rule and Venn Diagrams
When we want to find the probability that either event \(A\) occurs, event \(B\) occurs, or both occur, we use the Addition Rule.
The General Addition Rule
For any two events \(A\) and \(B\):
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
Why do we subtract \(P(A \cap B)\)?
Imagine you count all the people who like tea, and then add all the people who like coffee. People who like both have just been counted twice! We subtract \(P(A \cap B)\) once to fix the double count.
Mutually Exclusive Events
Two events are mutually exclusive if they cannot happen at the same time. For example, rolling a \(3\) and rolling a \(5\) on a single die roll.
• If \(A\) and \(B\) are mutually exclusive: \(P(A \cap B) = 0\)
• The Addition Rule simplifies to: \(P(A \cup B) = P(A) + P(B)\)
Step-by-Step Example: Venn Diagrams
Question: In a group of \(100\) students, \(60\) study Biology (\(B\)), \(45\) study Chemistry (\(C\)), and \(20\) study both. A student is chosen at random. Find \(P(B \cup C)\) and the probability that the student studies only Biology.
Step 1: Fill in the intersection first.
Number of students studying both: \(20\), so \(P(B \cap C) = \frac{20}{100} = 0.20\).
Step 2: Find the "only" regions.
• Biology only: \(60 - 20 = 40\) students \(\implies P(\text{Biology only}) = \frac{40}{100} = 0.40\)
• Chemistry only: \(45 - 20 = 25\) students \(\implies P(\text{Chemistry only}) = \frac{25}{100} = 0.25\)
Step 3: Calculate \(P(B \cup C)\).
Using the formula: \(P(B \cup C) = P(B) + P(C) - P(B \cap C) = \frac{60}{100} + \frac{45}{100} - \frac{20}{100} = \frac{85}{100} = 0.85\).
Notice that adding the separate regions gives the same result: \(0.40 + 0.20 + 0.25 = 0.85\).
Key Takeaway: When drawing Venn diagrams, always start by writing the probability or frequency in the central overlapping section (\(A \cap B\)), then work outwards!
3. Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. This "given that" condition reduces our sample space.
The Formula for Conditional Probability
The probability of event \(A\) happening, given that event \(B\) has happened, is written as \(P(A|B)\):
\(P(A|B) = \frac{P(A \cap B)}{P(B)}\), where \(P(B) > 0\)
Similarly: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\), where \(P(A) > 0\)
Everyday Analogy: Imagine guessing if it will snow today (\(A\)). That might have a low probability. But what if you are told the temperature outside is \(-5^\circ\text{C}\) and skies are overcast (\(B\))? Given this new information, the probability \(P(A|B)\) becomes significantly higher!
The Multiplication Rule
By rearranging the conditional probability formula, we get the General Multiplication Rule:
\(P(A \cap B) = P(A|B) \times P(B)\) or \(P(A \cap B) = P(B|A) \times P(A)\)
4. Independent Events
Two events \(A\) and \(B\) are independent if the occurrence of one event has no effect on the probability of the other event occurring.
Mathematical Definitions of Independence
Events \(A\) and \(B\) are independent if and only if:
\(P(A|B) = P(A)\) (and \(P(B|A) = P(B)\))
Substituting this into our multiplication rule gives the formal test for independence:
\(P(A \cap B) = P(A) \times P(B)\)
Common Exam Trap: "Mutually Exclusive" vs "Independent"
Do not confuse these two terms—they mean entirely different things!
• Mutually Exclusive: Events cannot happen together (\(P(A \cap B) = 0\)). If one happens, the other cannot happen.
• Independent: Events do not influence each other (\(P(A \cap B) = P(A) \times P(B)\)). One happening gives no information about whether the other happens.
Note: If two non-zero events are mutually exclusive, they cannot be independent!
Testing for Independence: Worked Example
Question: Given that \(P(A) = 0.6\), \(P(B) = 0.5\), and \(P(A \cup B) = 0.8\), determine whether \(A\) and \(B\) are independent.
Step 1: Find \(P(A \cap B)\) using the Addition Rule.
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(0.8 = 0.6 + 0.5 - P(A \cap B)\)
\(0.8 = 1.1 - P(A \cap B)\)
\(P(A \cap B) = 1.1 - 0.8 = 0.3\)
Step 2: Calculate the product \(P(A) \times P(B)\).
\(P(A) \times P(B) = 0.6 \times 0.5 = 0.3\)
Step 3: Compare and conclude.
Since \(P(A \cap B) = P(A) \times P(B) = 0.3\), events \(A\) and \(B\) are independent.
Key Takeaway: In an exam, always show the numerical comparison: state \(P(A \cap B)\), state \(P(A) \times P(B)\), and write a clear conclusion showing whether they are equal.
5. Tree Diagrams and Sequential Events
Tree diagrams are brilliant for visualising multi-stage experiments, especially when working with conditional probability (such as drawing items without replacement).
Rules for Tree Diagrams:
1. Multiply along the branches to find the intersection of outcomes: \(P(A \text{ and then } B) = P(A) \times P(B|A)\).
2. Add the probabilities down different paths to find the total probability of an outcome that can happen in multiple ways.
3. The branches radiating from any single node must always add up to \(1\).
Worked Example: Sampling Without Replacement
Question: A bag contains \(5\) Red counters and \(3\) Blue counters (total = \(8\)). Two counters are drawn one after another without replacement. Find the probability that:
(a) Both counters are Red.
(b) The counters are of different colours.
Step-by-step Solution:
First Draw:
• \(P(R_1) = \frac{5}{8}\)
• \(P(B_1) = \frac{3}{8}\)
Second Draw (Given First Draw):
• If first was Red: \(P(R_2|R_1) = \frac{4}{7}\) and \(P(B_2|R_1) = \frac{3}{7}\)
• If first was Blue: \(P(R_2|B_1) = \frac{5}{7}\) and \(P(B_2|B_1) = \frac{2}{7}\)
Calculations:
(a) Both Red:
\(P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\)
(b) Different Colours (Red then Blue OR Blue then Red):
• \(P(R_1 \cap B_2) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
• \(P(B_1 \cap R_2) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)
• Total Probability = \(\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\)
Key Takeaway: When objects are not replaced, remember to decrease both the numerator and the denominator on the second branch!
6. Two-Way Contingency Tables
Two-way tables display categorical data clearly and make calculating conditional probabilities straightforward.
Worked Example:
A survey of \(120\) athletes recorded their sport and whether they experienced an injury this season:
• Athletics: \(18\) Injured, \(42\) Not Injured (Total = \(60\))
• Swimming: \(6\) Injured, \(54\) Not Injured (Total = \(60\))
• Total: \(24\) Injured, \(96\) Not Injured (Total = \(120\))
Let \(I\) be the event that an athlete is Injured, and \(A\) be the event that an athlete does Athletics.
1. Find \(P(I)\):
\(P(I) = \frac{\text{Total Injured}}{\text{Grand Total}} = \frac{24}{120} = 0.2\)
2. Find \(P(I|A)\) (Probability of injury given the athlete does athletics):
Look only along the Athletics row (where the new total is \(60\)):
\(P(I|A) = \frac{18}{60} = 0.3\)
3. Find \(P(A|I)\) (Probability of doing athletics given the athlete is injured):
Look only down the Injured column (where the new total is \(24\)):
\(P(A|I) = \frac{18}{24} = 0.75\)
Notice: \(P(I|A) \ne P(A|I)\). The condition changes the denominator!
7. Quick Review & Formula Summary
Keep these fundamental formulas at your fingertips for your CCEA AS 2 exam:
• Complement: \(P(A') = 1 - P(A)\)
• Addition Rule (General): \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
• Mutually Exclusive Events: \(P(A \cap B) = 0 \implies P(A \cup B) = P(A) + P(B)\)
• Conditional Probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
• Multiplication Rule: \(P(A \cap B) = P(A|B) \times P(B)\)
• Independent Events Test: \(P(A \cap B) = P(A) \times P(B)\) or \(P(A|B) = P(A)\)
Top Exam Tips:
1. Read carefully: Words like "given that", "if", or "knowing that" signal conditional probability.
2. Check your values: If you get a probability greater than \(1\) or a negative number, stop and recheck your working.
3. Show full substitution: Always write out the formula, show the numbers substituted in, and then state your final answer clearly as an exact fraction or decimal.