S1.1: Further Probability – Mastering the Rules
Welcome to Further Probability! If you’ve grasped the basics of calculating chances, this chapter is where we give you the essential tools—the mathematical laws—to handle more complicated scenarios involving multiple events happening together or one after the other. Don't worry if this seems tricky at first; we break down each rule with clear examples!
The Foundation: Events and Set Notation
In probability, we use specific notation to describe how events relate to each other. While using this notation isn't always essential for answering the questions, understanding it makes the rules much clearer!
Key Notation Review
- Event \(A\): A specific outcome (e.g., rolling a 6).
- \(P(A)\): The probability of event A occurring. Remember, \(0 \le P(A) \le 1\).
- \(A'\) (A prime): The Complement of event A. This means 'A does not happen'.
- \(A \cap B\) (A intersect B): This means Event A AND Event B occur. Think of this as the overlap.
- \(A \cup B\) (A union B): This means Event A OR Event B or both occur.
Analogy: Imagine two circles (A and B) representing events. The intersection (\(\cap\)) is where they overlap. The union (\(\cup\)) is everything contained within both circles.
Section 1: The Addition Law (Dealing with 'OR' Events)
The Addition Law helps us find the probability of one event OR another event happening. This is formally called finding the Union, \(P(A \cup B)\).
1. Mutually Exclusive Events (The Simple Rule)
Two events, A and B, are Mutually Exclusive if they cannot happen at the same time. There is no overlap, so \(P(A \cap B) = 0\).
The Formula (for two or more events):
$$P(A \cup B) = P(A) + P(B)$$
Example: If you roll a single die, the event of rolling a 2 (A) and the event of rolling a 5 (B) are mutually exclusive. You cannot roll both at once.
Key Takeaway: If the events cannot share outcomes, you simply add their probabilities.
2. The General Addition Law (When Events Overlap)
If two events A and B are not mutually exclusive (meaning they can happen at the same time, so \(P(A \cap B) > 0\)), we must use the General Addition Law to avoid double-counting the overlap.
The Formula (for two events only):
$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$
Why do we subtract the intersection?
When you add \(P(A)\) and \(P(B)\), you count the students who do BOTH activities twice! Subtracting \(P(A \cap B)\) once corrects this count.
Example: In a class, 40% of students like Math (M) and 50% like Physics (P). 20% like both (\(M \cap P\)).
What is the probability a student likes Math OR Physics (\(M \cup P\))?
$$P(M \cup P) = P(M) + P(P) – P(M \cap P)$$
$$P(M \cup P) = 0.40 + 0.50 – 0.20 = 0.70$$
Quick Review: Addition Law
The word OR means ADD (but watch out for overlap!)
- If No Overlap (Mutually Exclusive): Add simply.
- If Overlap: Add, then Subtract the overlap (the intersection).
Section 2: Complementary Events
Often, finding the probability of an event *not* happening is easier than calculating the event itself, especially if the event covers many outcomes (e.g., rolling "at least one six").
The probability of the complement, \(A'\), is:
$$P(A') = 1 – P(A)$$
Tip: If a problem asks for the probability of "at least one," using the complement rule is almost always the easiest method! You calculate 1 minus the probability of "none."
Section 3: The Multiplication Law (Dealing with 'AND' Events)
The Multiplication Law helps us find the probability of event A AND event B happening in sequence or simultaneously. This is formally called finding the Intersection, \(P(A \cap B)\).
Memory Aid: The word AND means MULTIPLY!
1. Conditional Probability
When the outcome of the first event (A) influences the probability of the second event (B), the events are Dependent.
We use the concept of Conditional Probability, written as \(P(B|A)\), which means "the probability of B happening, given that A has already happened."
The General Multiplication Formula (for two or more events):
$$P(A \cap B) = P(A) \times P(B|A)$$
This formula can also be rearranged to find the conditional probability:
$$P(B|A) = \frac{P(A \cap B)}{P(A)}$$
Example (Dependent): Drawing two cards from a deck without replacement.
A = Drawing a King first. \(P(A) = \frac{4}{52}\).
B = Drawing a King second.
Since the first King wasn't replaced, there are only 3 Kings left and 51 cards total. So, the probability of B GIVEN A occurred is \(P(B|A) = \frac{3}{51}\).
$$P(\text{2 Kings}) = P(A \cap B) = \frac{4}{52} \times \frac{3}{51}$$
Accessibility Check: Conditional Probability Explained
Conditional probability limits your focus to a new, smaller sample space. When calculating \(P(B|A)\), you are no longer looking at the whole world of possible outcomes, but only at the outcomes where A has already happened.
2. Independent Events
Two events, A and B, are Independent if the occurrence of A does not affect the probability of B.
If A and B are independent, then the conditional probability \(P(B|A)\) is simply equal to \(P(B)\).
The Multiplication Formula (for two or more independent events):
$$P(A \cap B) = P(A) \times P(B)$$
Example (Independent): Tossing a coin (Heads, H) and rolling a die (Six, S).
Tossing a coin does not change the chances of rolling a six.
$$P(H \cap S) = P(H) \times P(S) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$
🔥 Common Mistake Alert!
Students often confuse Mutually Exclusive events (which use the Addition Law) and Independent events (which use the Multiplication Law).
- Mutually Exclusive: Cannot happen at the same time (e.g., getting an A AND a B on the same test). Probability of intersection (\(P(A \cap B)\)) is ZERO.
- Independent: One does not affect the other (e.g., flipping a coin and the weather). Probability of intersection is \(P(A) \times P(B)\).
If two events have any probability greater than zero, they cannot be both Mutually Exclusive and Independent.
Section 4: Application of Probability Laws
The syllabus confirms that you will primarily solve problems using direct application of the laws, counting equally likely outcomes, or using frequency/relative frequency tables.
Did you know? Although the examiners will not require you to draw them, using tools like Tree Diagrams for dependent events and Venn Diagrams for unions and intersections can be incredibly helpful for visualizing the problem and checking your work!
Step-by-Step Example using a Frequency Table
Suppose a survey records 100 people's travel methods to work (Car or Bus) and their gender (Male or Female).
| Car (C) | Bus (B) | Total | |
|---|---|---|---|
| Male (M) | 30 | 10 | 40 |
| Female (F) | 45 | 15 | 60 |
| Total | 75 | 25 | 100 |
We can use the table to find relative frequencies (probabilities):
1. Simple Probability: What is \(P(M)\)? (Probability of being Male)
$$P(M) = \frac{40}{100} = 0.4$$
2. Intersection (AND): What is \(P(F \cap B)\)? (Female AND take the Bus)
$$P(F \cap B) = \frac{15}{100} = 0.15$$
3. Union (OR, General Addition Law): What is \(P(C \cup M)\)? (Car OR Male)
We use the formula: \(P(C \cup M) = P(C) + P(M) – P(C \cap M)\)
$$P(C) = \frac{75}{100} = 0.75$$
$$P(M) = \frac{40}{100} = 0.40$$
$$P(C \cap M) = \frac{30}{100} = 0.30$$
$$P(C \cup M) = 0.75 + 0.40 – 0.30 = 0.85$$
(Check: 30 + 10 + 45 = 85 people satisfy the condition)
4. Conditional Probability: What is \(P(C|M)\)? (Car, GIVEN they are Male)
We restrict our view only to the Male row (Total = 40).
$$P(C|M) = \frac{\text{Number of Males taking Car}}{\text{Total Males}} = \frac{30}{40} = 0.75$$
Alternatively, using the conditional formula:
$$P(C|M) = \frac{P(C \cap M)}{P(M)} = \frac{0.30}{0.40} = 0.75$$
Key Takeaway: Practice applying the laws directly, and always clearly identify whether events are independent, dependent, or mutually exclusive before choosing your formula.