If P(A) = 0.4, P(B) = 0.5, P(A∩B) = 0.2, find P(A|B).

Conditional Probability: P(A|B) Explained

01 — The Intuition: Why Knowledge Shrinks the Sample Space

Imagine rolling a fair six-sided die. Before it lands, every outcome {1, 2, 3, 4, 5, 6} has equal probability 1/6. Now someone peeks and tells you: "It landed on an even number." The moment you know that, you discard the odd outcomes. Your sample space shrinks from {1,2,3,4,5,6} to just {2,4,6}, and the probability that it was a 4 jumps from 1/6 to 1/3.

That is the entire idea behind conditional probability, you condition on partial information, which restricts the sample space and rescales all the probabilities inside it.

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Key Insight: P(A | B) asks: "Given that B has already occurred, what is the probability of A?" You are no longer looking at the full sample space — only the slice where B is true.

02 — The Formal Definition

Let A and B be events in a sample space Ω with P(B) > 0. The conditional probability of A given B is:

Conditional Probability — Core Formula

P(A | B) = P(A ∩ B) / P(B)

Read: "the probability of A given B equals the probability of A-and-B divided by the probability of B."

Why does this formula make sense? Think of it geometrically. P(B) measures the size of event B. P(A ∩ B) measures the portion of B that is also inside A. The ratio tells you what fraction of the B-world is also in the A-world.

The Requirement P(B) > 0

Conditioning on an impossible event is undefined: you cannot "know" something that can never happen. Always check P(B) > 0 before applying the formula. This is the single most common oversight in exam questions.

03 — Visualising with a Venn Diagram

Venn Diagram, P(A | B) focuses only inside B

P(A | B) = area of red intersection ÷ area of gold ellipse B. We zoom in on B and measure what fraction of it overlaps with A.

04 — The Multiplication Rule

Rearranging the definition of conditional probability gives the multiplication rule, which lets you compute the probability of two events both occurring:

Multiplication Rule

P(A ∩ B) = P(A | B) · P(B)

Equivalently: P(A ∩ B) = P(B | A) · P(A) — both are valid because A ∩ B = B ∩ A.

This is the engine behind tree diagrams. At each branch you write a conditional probability; multiply along the path to get the joint probability at the leaf.

05 — Tree Diagrams: Laying Out Sequential Events

A tree diagram organises sequential events so that multiplying along any path gives the probability of that full sequence. Each set of branches from a node must sum to 1 (they're exhaustive and mutually exclusive, given what came before).

Tree Diagram: Drawing Two Cards Without Replacement (Simplified)

A bag has 3 red and 3 black balls. Drawing two without replacement: multiply along each path to get the joint probability. All four leaves sum to 30/30 = 1. ✓

06 — Statistical Independence

Two events A and B are independent if knowing B occurred gives you no information about A. Formally:

Independence Condition

P(A | B) = P(A)

Equivalently: P(A ∩ B) = P(A) · P(B) — this is the most useful form for calculations.

Independence is a symmetric relation: if A is independent of B, then B is independent of A. Independence is also not the same as mutually exclusive, in fact, if A and B are mutually exclusive (P(A ∩ B) = 0) and both have positive probability, they are necessarily dependent, since knowing B rules out A entirely.

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Common Mistake: "Mutually exclusive" and "independent" are opposite extremes, not synonyms. Mutually exclusive events maximally affect each other — learning that one occurred makes the other impossible (P = 0).

07 — Ten Worked Examples

EX 01 Basic conditional probability from a table

In a class of 40 students: 18 study Maths, 14 study Physics, 8 study both. One student is chosen at random. Given they study Maths, what is the probability they also study Physics?

Identify: A = studies Physics, B = studies Maths.
P(B) = 18/40. P(A ∩ B) = 8/40.

Apply the formula: P(A|B) = P(A∩B)/P(B) = (8/40)/(18/40)

Simplify: = 8/18 = 4/9

P(Physics | Maths) = 4/9 ≈ 0.444 — about 44.4% of Maths students also study Physics.

EX 02 Using the multiplication rule

P(A) = 0.5, P(B|A) = 0.6. Find P(A ∩ B).

Rearrange the definition: P(A ∩ B) = P(B|A) · P(A)

P(A ∩ B) = 0.6 × 0.5 = 0.30

P(A ∩ B) = 0.30

EX 03 Checking independence

P(A) = 0.3, P(B) = 0.4, P(A ∩ B) = 0.12. Are A and B independent?

For independence: P(A ∩ B) should equal P(A) · P(B).

P(A) · P(B) = 0.3 × 0.4 = 0.12

0.12 = 0.12 ✓ — the condition holds.

A and B are independent.

EX 04 Deck of cards — without replacement

Two cards drawn without replacement from a standard 52-card deck. What is the probability both are aces?

P(first ace) = 4/52 = 1/13

P(second ace | first ace) = 3/51 = 1/17 — one ace removed, deck has 51 cards.

P(both aces) = (4/52) × (3/51) = 12/2652 = 1/221

P(AA) = 1/221 ≈ 0.00452

EX 05 Medical test — false positive scenario

A disease affects 2% of a population. A test is 95% accurate for those who have the disease and 90% accurate for those who don't (i.e., 10% false positive rate). A patient tests positive. What is P(disease | positive test)?

P(D) = 0.02, P(D') = 0.98. P(+|D) = 0.95, P(+|D') = 0.10.

Law of total probability: P(+) = P(+|D)·P(D) + P(+|D')·P(D')
= 0.95×0.02 + 0.10×0.98 = 0.019 + 0.098 = 0.117

P(D|+) = P(+|D)·P(D) / P(+) = 0.019/0.117

P(D | positive) ≈ 0.162 — only 16.2% despite a 95% accurate test! — Base rate matters enormously.

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Note: Example 05 is the starting point for Bayes' Theorem — the systematic method for updating probabilities with new evidence. Follow that link to go deeper.

EX 06 Conditional probability from first principles (die roll)

A fair die is rolled. Given the result is greater than 3, what is the probability it is even?

B = {4, 5, 6} (greater than 3). P(B) = 3/6 = 1/2.

A = {2, 4, 6} (even). A ∩ B = {4, 6}. P(A ∩ B) = 2/6 = 1/3.

P(A|B) = (1/3)/(1/2) = 2/3

P(even | > 3) = 2/3 — from {4,5,6}, two of three are even.

EX 07 Tree diagram with three outcomes

A bag has 4 red and 6 blue balls. Two balls are drawn without replacement. Find P(second ball red | first ball blue).

After drawing a blue ball, 9 balls remain: 4 red and 5 blue.

P(second red | first blue) = 4/9 directly from the updated bag.

P(R₂ | B₁) = 4/9 ≈ 0.444

EX 08 Verifying non-independence

P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.25. Are A and B independent?

Test: P(A) · P(B) = 0.4 × 0.5 = 0.20

P(A ∩ B) = 0.25 ≠ 0.20

P(A|B) = 0.25/0.5 = 0.50 ≠ P(A) = 0.40 — confirms dependence.

A and B are dependent. Knowing B increases the probability of A.

EX 09 Law of Total Probability

Factory line 1 produces 60% of output with a 3% defect rate. Line 2 produces 40% with a 5% defect rate. An item is selected at random. What is the probability it is defective?

P(D) = P(D|L₁)·P(L₁) + P(D|L₂)·P(L₂)

= 0.03 × 0.60 + 0.05 × 0.40

= 0.018 + 0.020 = 0.038

P(defective) = 0.038 = 3.8%

EX 10 Conditional probability with complements

P(A|B) = 0.7. Find P(A'|B).

Inside the event B, all conditional probabilities must sum to 1.

P(A|B) + P(A'|B) = 1

P(A'|B) = 1 − 0.7 = 0.3

P(A' | B) = 0.3 — the complement rule works within any fixed conditioning event.

08 — Common Mistakes

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Mistake 1 — Reversing the conditioning: P(A|B) ≠ P(B|A) in general. "The probability of disease given a positive test" is not the same as "the probability of a positive test given disease." Confusing these is called the prosecutor's fallacy.

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Mistake 2 — Forgetting to update the denominator: In "without replacement" problems, the denominator shrinks after each draw. Always update the sample space size before writing the next conditional probability.

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Mistake 3 — Assuming independence without justification: Never assume P(A ∩ B) = P(A)·P(B) unless the problem explicitly states independence, or you can logically justify it from the physical setup.

🔗 Related Topics in Probability & Statistics

Basic Probability Bayes' Theorem Discrete Distributions Binomial Distribution Poisson Distribution Normal Distribution Expected Value & Variance Hypothesis Testing All Probability Topics

Conditional Probability
P(A | B) — What Changes When You Know More

01 — The Intuition: Why Knowledge Shrinks the Sample Space

02 — The Formal Definition

The Requirement P(B) > 0

03 — Visualising with a Venn Diagram

04 — The Multiplication Rule

05 — Tree Diagrams: Laying Out Sequential Events

06 — Statistical Independence

07 — Ten Worked Examples

08 — Common Mistakes

🔗 Related Topics in Probability & Statistics

09 — Quiz: Conditional Probability (10 Questions)