More about Probability

Master conditional probability, addition laws, and multi-stage tree diagrams. Learn how to construct mathematical proofs of independence and avoid costly careless errors.

1. The Core Concept

Probability as Ratios: At its core, probability is about comparing favorable outcomes to total outcomes. However, in advanced scenarios, the "total outcomes" (sample space) changes dynamically — either because an event has already occurred (Conditional Probability) or because objects are removed from the system (Without Replacement).

2. DSE More about Probability Cheat Sheet — 5 Key Topics

Scan this compact reference table to review every essential concept, step, and shortcut in under 10 seconds.

TopicKey Concept & StepsExam Shortcuts & Tips
1. Addition LawP(A ∪ B) = P(A) + P(B) − P(A ∩ B)• If A and B are mutually exclusive:
P(A ∩ B) = 0 ⟹ P(A ∪ B) = P(A) + P(B)
2. Conditional ProbabilityP(A | B) = P(A ∩ B) / P(B)• Read as: "Probability of A given B has occurred."
• The denominator shrinks to only include the outcomes inside event B.
3. Independence of EventsA and B are independent if and only if:
P(A ∩ B) = P(A) × P(B)
(equivalently, P(A | B) = P(A))
Proof Requirement:
You must calculate the left-hand side and right-hand side separately, then write whether they are equal.
4. Tree Diagrams (AP/GP)Along branches (AND): Multiply.
Across branches (OR): Add.
Used for multi-stage selection. Perfect for tracking "without replacement" games where probabilities change.
5. P&C ProbabilityP(Event) = Favorable C(n,r) or P(n,r) / Total C(n,r) or P(n,r)Used for drawing balls, cards, or forming committees where selection order does or does not matter.

💡 Exam Traps & Secrets

⚠️ Conditional P(A|B) vs. Intersection P(A∩B) TrapThe difference: P(A ∩ B) calculates the probability of both A and B occurring out of the entire original group. P(A | B) restricts the world to only group B, and calculates the probability of A occurring within that group.
⚠️ The Independence Proof Trap (Paper 1 Step Marks)Never assume two events are independent just because they "sound" independent in the question text. You must prove it algebraically. How to write your steps:
• Calculate LHS = P(A ∩ B)
• Calculate RHS = P(A) × P(B)
• Compare: "Since LHS = RHS (or LHS ≠ RHS), the events A and B are independent (or dependent)."
⚠️ "Without Replacement" Denominator DecrementIn multi-stage drawing questions, if items are drawn successively without replacement, you must remember to decrease both your numerator and denominator by 1 for the subsequent branches. Example: Drawing 2 red balls from a bag of 4 red and 6 blue (total 10):
P(Both Red) = 4/10 × 3/9 = 12/90 = 2/15

✏️ Self-Test: Quick Interactive Practice

How to study: Click any question below to instantly load it on the coordinate grid and check your steps.

(Tree diagram calculations with changing denominators)A bag has 4 red and 6 blue balls. Two balls are drawn successively without replacement. Find P(both are red) and P(at least one is red).
(Addition law, conditional probability, and independence proof steps)Given P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2. Find P(A ∪ B) and P(A | B). Are A and B independent?
(Combination-based probability calculation)From a class of 20 students (12 boys and 8 girls), a committee of 3 is chosen. Find the probability that there are at least 2 girls in the committee.

Frequently Asked Questions

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