Conditional Probability

Calculate conditional probabilities for GCSE Maths. Use tree diagrams and two-way tables for dependent events.

Conditional probability is the probability of an event given that another event has already occurred. It's a Higher GCSE topic that appears in tree diagrams and two-way tables.

Core Concepts

Notation

P(AB)P(A|B) = probability of A given B has occurred.

Formula

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

From Two-Way Tables

Pass Fail Total
Studied 35 5 40
Didn't 10 15 25
Total 45 20 65

P(PassStudied)=3540=0.875P(\text{Pass} | \text{Studied}) = \frac{35}{40} = 0.875

P(StudiedPass)=35450.778P(\text{Studied} | \text{Pass}) = \frac{35}{45} \approx 0.778

From Tree Diagrams

Conditional probability appears naturally — the second branches already show conditional probabilities.

Independent Events

If P(AB)=P(A)P(A|B) = P(A), events A and B are independent.

For independent events: P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).

Worked Example: Example 1

Problem

P(A)=0.4P(A) = 0.4, P(B)=0.5P(B) = 0.5, P(AB)=0.15P(A \cap B) = 0.15.

P(AB)=0.150.5=0.3P(A|B) = \frac{0.15}{0.5} = 0.3.

Since 0.30.4=P(A)0.3 \neq 0.4 = P(A), events are not independent.

Solution

Worked Example: Example 2

Problem

Bag: 4 red, 6 blue. Pick two without replacement.

P(2nd red1st red)=39=13P(\text{2nd red} | \text{1st red}) = \frac{3}{9} = \frac{1}{3}

Solution

Practice Problems

    1. From the two-way table above, find P(FailDidn’t study)P(\text{Fail} | \text{Didn't study}).
    1. Are events A and B independent if P(A)=0.3P(A) = 0.3, P(B)=0.6P(B) = 0.6, P(AB)=0.18P(A \cap B) = 0.18?

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Key Takeaways

  • P(AB)=P(AB)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}.

  • Two-way tables: restrict to the given row/column.

  • Independent: P(AB)=P(A)P(A|B) = P(A)P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).

  • Tree diagram second branches are already conditional probabilities.

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