Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: Probability | Statistics & Logic
🔹 Introduction
Conditional probability tells us how likely an event is, given that another event has already occurred.
It’s one of the most powerful tools in probability — and one of the most misunderstood.
Conditional probability connects what we know with what we want to find out.
We’ll understand it step-by-step — visually using Venn Diagrams and Tree Diagrams.
🧭 1️⃣ Definition of Conditional Probability
For any two events AAA and BBB (with P(B)>0P(B) > 0P(B)>0): P(A∣B)=P(A∩B)P(B)\boxed{P(A|B) = \frac{P(A \cap B)}{P(B)}}P(A∣B)=P(B)P(A∩B)
- P(A∣B)P(A|B)P(A∣B): Probability that A occurs given B has occurred.
- P(A∩B)P(A \cap B)P(A∩B): Probability that both A and B occur.
- P(B)P(B)P(B): Probability that B occurs.
✅ Think of it as “narrowing the sample space” to only cases where B happens.
🔹 Example 1
Suppose 60% of students play sports (S), and 30% play both sports and music (M).
Find P(M∣S)P(M|S)P(M∣S): probability that a student plays music given that they play sports. P(S)=0.6,P(M∩S)=0.3P(S) = 0.6, \quad P(M \cap S) = 0.3P(S)=0.6,P(M∩S)=0.3 P(M∣S)=P(M∩S)P(S)=0.30.6=0.5P(M|S) = \frac{P(M \cap S)}{P(S)} = \frac{0.3}{0.6} = 0.5P(M∣S)=P(S)P(M∩S)=0.60.3=0.5
✅ So half of all sports players also play music.
⚡️ 2️⃣ Visualizing with a Venn Diagram
(Illustration: Two overlapping circles, A and B, with shaded intersection.)
In the Venn diagram:
- The rectangle represents the entire sample space.
- A∩BA \cap BA∩B is the overlapping region — both events occur.
- P(A∣B)P(A|B)P(A∣B) focuses only on circle B — then asks, what fraction of B overlaps with A?
✅ Geometrically: P(A∣B)=Area of (A∩B)Area of BP(A|B) = \frac{\text{Area of } (A \cap B)}{\text{Area of } B}P(A∣B)=Area of BArea of (A∩B)
🔹 Example 2 — Venn Logic
In a class:
- 40% study Math (M),
- 30% study Physics (P),
- 20% study both.
Find P(M∣P)P(M|P)P(M∣P) and P(P∣M)P(P|M)P(P∣M). P(M∣P)=P(M∩P)P(P)=0.20.3=23P(M|P) = \frac{P(M \cap P)}{P(P)} = \frac{0.2}{0.3} = \frac{2}{3}P(M∣P)=P(P)P(M∩P)=0.30.2=32 P(P∣M)=P(M∩P)P(M)=0.20.4=0.5P(P|M) = \frac{P(M \cap P)}{P(M)} = \frac{0.2}{0.4} = 0.5P(P∣M)=P(M)P(M∩P)=0.40.2=0.5
✅ Order matters — conditional probabilities are not symmetric.
🌳 3️⃣ Conditional Probability Using Tree Diagrams
A tree diagram helps visualize sequences of events, especially when they occur in stages.
Each branch represents an event and its conditional probability.
🔹 Example 3 — Tree Diagram for Two Events
A bag has:
- 3 Red balls (R)
- 2 Blue balls (B)
One ball is drawn, replaced, and drawn again.
(Illustration: Tree diagram with branches for R/B on both draws.)
| Path | Probability |
|---|---|
| R then R | 35×35=925\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}53×53=259 |
| R then B | 35×25=625\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}53×52=256 |
| B then R | 25×35=625\frac{2}{5} \times \frac{3}{5} = \frac{6}{25}52×53=256 |
| B then B | 25×25=425\frac{2}{5} \times \frac{2}{5} = \frac{4}{25}52×52=254 |
✅ Sum = 1.
✅ Each branch = conditional multiplication: P(A∩B)=P(A)×P(B∣A)P(A \cap B) = P(A) \times P(B|A)P(A∩B)=P(A)×P(B∣A)
🔹 Conditional Probability from Tree
If we know a Blue ball was drawn first, find P(2nd Red∣1st Blue)P(\text{2nd Red}|\text{1st Blue})P(2nd Red∣1st Blue).
From tree: P(2nd Red∣1st Blue)=35=0.6P(\text{2nd Red}|\text{1st Blue}) = \frac{3}{5} = 0.6P(2nd Red∣1st Blue)=53=0.6
✅ The second draw is independent — conditional probability = unconditional.
🔹 Example 4 — Dependent Case (Without Replacement)
Same bag (3R, 2B), but no replacement.
| Path | Probability |
|---|---|
| R then R | 35×24=310\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}53×42=103 |
| R then B | 35×24=310\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}53×42=103 |
| B then R | 25×34=310\frac{2}{5} \times \frac{3}{4} = \frac{3}{10}52×43=103 |
| B then B | 25×14=110\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}52×41=101 |
✅ Here, second draw depends on the first → conditional probabilities differ.
🧩 4️⃣ The Multiplication Rule
P(A∩B)=P(A)P(B∣A)=P(B)P(A∣B)\boxed{P(A \cap B) = P(A)P(B|A) = P(B)P(A|B)}P(A∩B)=P(A)P(B∣A)=P(B)P(A∣B)
This rule links joint and conditional probabilities.
Used to compute complex paths in tree diagrams.
🔹 Example 5 — Application
If P(A)=0.6,P(B∣A)=0.5P(A) = 0.6, P(B|A) = 0.5P(A)=0.6,P(B∣A)=0.5,
then P(A∩B)=0.6×0.5=0.3P(A \cap B) = 0.6 \times 0.5 = 0.3P(A∩B)=0.6×0.5=0.3.
If P(B)=0.4P(B) = 0.4P(B)=0.4,
then P(A∣B)=P(A∩B)P(B)=0.30.4=0.75P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.4} = 0.75P(A∣B)=P(B)P(A∩B)=0.40.3=0.75.
✅ P(A∣B)≠P(B∣A)P(A|B) \neq P(B|A)P(A∣B)=P(B∣A). Order matters.
🎯 5️⃣ Conditional Probability Table
| Event | P(A)P(A)P(A) | P(B∣A)P(B|A)P(B∣A) | P(A∩B)=P(A)P(B∣A)P(A \cap B) = P(A)P(B|A)P(A∩B)=P(A)P(B∣A) |
|——–|————-|—————|——————————–|
| Example | 0.6 | 0.5 | 0.3 |
🧠 6️⃣ Independent vs Dependent Events
| Relationship | Definition | Formula |
|---|---|---|
| Independent | One does not affect the other | ( P(A |
| Dependent | One affects probability of other | ( P(A |
✅ Independence check: P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B)P(A∩B)=P(A)P(B)
📘 7️⃣ Real-Life Examples
- Medicine: P(Test Positive | Has Disease)
- Marketing: P(Buy Product | Clicked Ad)
- Weather: P(Rain | Clouds)
- Quality Control: P(Defective | From Factory B)
Conditional probability helps us update beliefs based on new information — the foundation of Bayesian reasoning.
🔹 Common Mistakes
- ❌ Confusing P(A∣B)P(A|B)P(A∣B) with P(B∣A)P(B|A)P(B∣A)
- ❌ Forgetting to divide by P(B)P(B)P(B) in formula
- ❌ Ignoring dependency in tree diagrams
- ❌ Adding probabilities instead of multiplying along branches
🌟 Why It Matters
Conditional probability lets us reason logically about uncertainty.
It’s used everywhere — from diagnosing diseases to predicting rain.
“Probability without condition is prediction.
Conditional probability is understanding.”
📘 Learn Beyond the Formula
At Math By Rishabh, we don’t just teach probability — we train reasoning under uncertainty.
In the Mathematics Elevate Mentorship Program, you’ll:
✅ Master conditional logic via Venn and tree diagrams,
✅ Apply to IB / A Level / AP problems,
✅ Learn real-world uses in data and AI.
🚀 Visualize uncertainty, reason with clarity.
👉 Book your personalized mentorship session now at MathByRishabh.com
