CAIE A-Level · Mathematics 9709 · Probability

Conditional Probability (PS1 – 5.3)

8 min readSyllabus 5.3PreviewBy Uzair Khan

Syllabus objective

Calculate and use conditional probabilities in simple cases, e.g. situations that can be represented by a sample space of equiprobable elementary events, or a tree diagram. The use of P(A | B) = P(A ∩ B) / P(B) may be required in simple cases.

Introduction

Conditional probability answers the question: given that one event has already occurred, how does that change the probability of another event? This idea appears throughout Probability & Statistics 1 exam papers — in questions involving two-way tables, tree diagrams, and explicit use of the formula P(AB)=P(AB)P(B)P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}. Mastering conditional probability is essential for the 9709 exam because it connects sample spaces, tree diagrams, and the multiplication rule into a single coherent framework.


Core Concept

What "conditional" means

The notation P(AB)P(A \mid B) is read as "the probability of AA given BB". It represents the probability that event AA occurs, under the condition that event BB is known (or assumed) to have occurred. Conditioning effectively restricts the sample space to only those outcomes in BB, and asks what fraction of those outcomes also lie in AA.

Intuition with a sample space: Suppose you roll two fair dice and record the outcomes as ordered pairs. There are 36 equiprobable elementary events. If you are told that the sum is at least 10, the sample space shrinks to {(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\} — just 6 outcomes. The conditional probability of "both dice show the same number given sum 10\geq 10" is 16\frac{1}{6} (only (5,5)(5,5) qualifies out of those 6).

The formula

P(AB)=P(AB)P(B),P(B)>0P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0

This formula is always valid. It is especially useful when you cannot list all elementary events but you do know (or can calculate) P(AB)P(A \cap B) and P(B)P(B).

Rearranging gives the Multiplication Rule (a prerequisite topic):

P(AB)=P(B)×P(AB)P(A \cap B) = P(B) \times P(A \mid B)

which is exactly what each branch-pair on a tree diagram computes.

Independence as a special case

Events AA and BB are independent if and only if P(AB)=P(A)P(A \mid B) = P(A). Knowing BB has occurred gives no information about AA. This is consistent with the multiplication rule for independent events: P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).


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