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Class 12 Statistics Notes · GSEB

Conditional and Multiplication

Probability — learn conditional probability P(A|B), independent events, and the multiplication rule with tree diagrams. GSEB Class 12 Statistics notes.

Last updated: 25 Aug 2026

Notes

Conditional Probability and Multiplication Rule

Conditional probability asks: "Given that B has happened, what is the probability of A?" This changes the sample space from S to B.

Conditional Probability

Conditional Probability — P(A|B)
The probability of event A occurring given that event B has already occurred. We restrict the sample space to B, then count how much of A falls inside B.

Conditional Probability Formula

P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

Samosa Stall Test

In your class of 40 students, 25 play cricket (C) and 15 play football (F), with 8 playing both. P(cricket | football) = P(C ∩ F)/P(F) = 8/40 ÷ 15/40 = 8/15 ≈ 0.53. Given that a student plays football, there's a 53% chance they also play cricket.

Independent Events

Independent Events
Two events A and B are independent if the occurrence of one does NOT affect the probability of the other. Mathematically: P(A|B) = P(A), which leads to the multiplication rule for independent events.

Multiplication Rule (Independent Events)

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)
AspectDependentIndependent
P(A|B) vs P(A)P(A|B) ≠ P(A)P(A|B) = P(A)
MultiplicationP(A∩B) = P(A)·P(B|A)P(A∩B) = P(A)·P(B)
ExampleDrawing cards without replacementTossing two coins

Common mistake

"Two dice are rolled. Are the events 'first die shows 3' and 'sum is 7' independent?" Check: P(A) = 1/6, P(B) = 6/36 = 1/6, P(A∩B) = 1/6 × 1/6 = 1/36? No — P(A∩B) = 1/6 (outcomes: 3,1; 3,2; 3,3; 3,4; 3,5; 3,6 — only 3,4 gives sum 7). So P(A∩B) = 1/36 = P(A)·P(B). They ARE independent!

General Multiplication Rule

General Multiplication Rule

P(A∩B)=P(A)×P(B∣A)=P(B)×P(A∣B)P(A \cap B) = P(A) \times P(B|A) = P(B) \times P(A|B)

Solved Example

Problem

A bag contains 5 red and 3 blue balls. Two balls are drawn one after another WITHOUT replacement. Find P(both red).

Solution

P(both red) = 5/14 ≈ 0.357

Explore: Probability Tree Diagram

Walk through the 5 steps to build a probability tree. Each step reveals more of the diagram.

Probability Tree Builder

Scenario: It rains 30% of days. If it rains, you carry an umbrella 90% of the time. If no rain, you carry one 20% of the time.

StartP = 0.3P = 0.7RNR0.90.10.20.8R∩U = 0.27R∩U' = 0.03R'∩U = 0.14R'∩U' = 0.56

Step 1: Define Event A

Rain (P = 0.3) vs No Rain (P = 0.7)

Solved Examples

Solved Example

Problem

P(A) = 0.6, P(B) = 0.4, P(A ∩ B) = 0.2. Find P(A|B) and P(B|A).

Solution

P(A|B) = 0.5, P(B|A) = 1/3

Solved Example

Problem

Two cards are drawn from a 52-card deck WITHOUT replacement. Find P(both are kings).

Solution

P(both kings) = 1/221 ≈ 0.0045

Key Takeaways

Key Takeaways

  • P(A|B) = P(A ∩ B)/P(B) — restricts sample space to B.
  • Independent events: P(A ∩ B) = P(A) × P(B) — one does not affect the other.
  • General multiplication: P(A ∩ B) = P(A) × P(B|A) — works always.
  • Without replacement → dependent events. With replacement → independent events.
  • Tree diagrams are powerful: multiply along branches, add across branches.

Practice

  1. P(A) = 0.5, P(B) = 0.3, P(A ∩ B) = 0.15. Are A and B independent?
  2. A bag has 4 white and 6 black balls. Two balls are drawn without replacement. Find P(both black).
  3. P(rain today) = 0.4. P(carrying umbrella | rain) = 0.8. P(carrying umbrella | no rain) = 0.1. Find P(carrying umbrella).

Explore Further