Conditional Probability

This lesson covers conditional probability, $P(A\mid B)$ , using tree diagrams and two-way tables to calculate conditional probabilities, and problems involving dependent events. This is primarily a Higher tier topic on the Edexcel GCSE Mathematics (1MA1) specification, though Foundation students encounter basic conditional ideas through "without replacement" questions.

What Is Conditional Probability?

Conditional probability is the probability of an event occurring given that another event has already occurred.

The notation $P(A\mid B)$ means "the probability of A given B" — the probability that event A happens, given that we know event B has happened.

$P(A\mid B) = \frac{P(A \cap B)}{P(B)}$

In words: the probability of A given B equals the probability of both A and B divided by the probability of B.

Which Tool Should I Use?

Before diving into a probability question, decide which representation fits best:

graph TD
    Q["Read the question"] --> Stages{"More than one stage<br/>(pick 1, pick 2, …)?"}
    Stages -->|Yes| Tree["Use a TREE DIAGRAM<br/>(multiply along, add between)"]
    Stages -->|No| Sets{"Classified by sets<br/>(e.g. A and B, overlapping)?"}
    Sets -->|Yes — 2 or 3 sets| Venn["Use a VENN DIAGRAM<br/>(intersection, union, complement)"]
    Sets -->|No — two categories| Table["Use a TWO-WAY TABLE<br/>(restrict to row/column for P(A|B))"]
    Sets -->|No — one event| Sample["Use a SAMPLE SPACE<br/>(list all equally likely outcomes)"]

    style Tree fill:#2980b9,color:#fff
    style Venn fill:#27ae60,color:#fff
    style Table fill:#8e44ad,color:#fff
    style Sample fill:#e67e22,color:#fff

Conditional probability questions can appear from any of these representations — but the formula $P(A\mid B) = \frac{P(A \cap B)}{P(B)}$ is the same every time.

Key Vocabulary

Term	Definition
Conditional probability	Probability of an event given that another event has occurred
P(A given B) or $P(A\mid B)$	Probability of A, knowing B has happened
Independent events	Events where $P(A\mid B) = P(A)$ — knowing B does not affect A
Dependent events	Events where $P(A\mid B)$ ≠ $P(A)$ — knowing B changes the probability of A

Conditional Probability from Two-Way Tables

Two-way tables are the most straightforward way to calculate conditional probability. The key insight is that "given B" restricts the sample space to only those in group B.

Worked Example 1

A school surveys 200 students about whether they walk or take the bus, and whether they are in Year 10 or Year 11.

	Walk	Bus	Total
Year 10	50	40	90
Year 11	30	80	110
Total	80	120	200

Find: (a) P(Walk) (b) P(Walk | Year 10) — the probability a student walks, given they are in Year 10 (c) P(Year 11 | Bus) — the probability a student is Year 11, given they take the bus

Solution:

(a) P(Walk) = 80/200 = 2/5

(b) Given Year 10: restrict to the 90 Year 10 students. Of these, 50 walk. P(Walk | Year 10) = 50/90 = 5/9

Key Point: For conditional probability from a table, the denominator is the total for the given condition (row or column total), not the grand total.

Conditional Probability from Tree Diagrams

Worked Example 2

A bag contains 5 red and 3 blue counters. A counter is drawn, not replaced, and a second counter is drawn. Find:

(a) P(second is red | first was blue) (b) P(first was red | second is red)

Tree Diagram:

First draw:

$P(R_1) = \frac{5}{8}$ , $P(B_1) = \frac{3}{8}$

Second draw after $R_1$ (4R, 3B remain, total 7):

$P(R_2) = \frac{4}{7}$ , $P(B_2) = \frac{3}{7}$

Second draw after $B_1$ (5R, 2B remain, total 7):

$P(R_2) = \frac{5}{7}$ , $P(B_2) = \frac{2}{7}$

Solution:

(a) Given the first was blue, we are on the $B_1$ branch. The second-draw probabilities are: $P(R_2 \mid B_1)$ = 5/7

This can be read directly from the tree diagram.

(b) This requires working backwards. We want $P(R_1 \mid R_2)$ .

Using the formula: $P(R_1 \mid R_2) = \frac{P(R_1 \cap R_2)}{P(R_2)}$

$P(R_1 \cap R_2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$

$P(R_2) = P(R_1 \cap R_2) + P(B_1 \cap R_2) = \frac{20}{56}$ + ( $\frac{3}{8} \times \frac{5}{7}$ ) = 20/56 + 15/56 = 35/56

$P(R_1 \mid R_2)$ = (20/56) / (35/56) = 20/35 = 4/7

The Conditional Probability Formula

$P(A\mid B) = \frac{P(A \cap B)}{P(B)}$

Rearranging: $P(A \cap B) = P(A\mid B) × P(B)$

This is the multiplication rule for dependent events.

Worked Example 3

$P(A)$ = 0.6, $P(B)$ = 0.5, $P(A\mid B)$ = 0.4. Find $P(A \cap B)$ and $P(A \cup B)$ .

Solution:

$P(A \cap B) = P(A\mid B) × P(B)$ = 0.4 × 0.5 = 0.2

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ = 0.6 + 0.5 − 0.2 = 0.9

Testing for Independence Using Conditional Probability

Two events A and B are independent if and only if:

$P(A\mid B) = P(A)$

This means knowing that B has occurred does not change the probability of A.

Worked Example 4

Using the data from Worked Example 1: P(Walk) = 80/200 = 0.4 P(Walk | Year 10) = 50/90 ≈ 0.556

Since P(Walk) ≠ P(Walk | Year 10), the events "walks" and "Year 10" are not independent. Year 10 students are more likely to walk than the overall student population.

Worked Example 5

100 people are surveyed. The results are:

	Likes chocolate	Does not like chocolate	Total
Male	36	24	60
Female	24	16	40
Total	60	40	100

P(likes chocolate) = 60/100 = 0.6 P(likes chocolate | male) = 36/60 = 0.6

Since these are equal, "likes chocolate" and "male" are independent events.

Harder Conditional Probability Problems (Higher)

Worked Example 6

A bag contains 8 balls: 5 are green and 3 are yellow. Two balls are drawn without replacement.

Given that at least one ball is green, find the probability that both balls are green.

Solution:

P(both green) = $\frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$ = 5/14

P(at least one green) = 1 − P(no green) = 1 − P(both yellow) = 1 − ( $\frac{3}{8} \times \frac{2}{7}$ ) = 1 − 6/56 = 50/56 = 25/28

P(both green | at least one green) = P(both green) / P(at least one green) = (5/14) / (25/28) = (5/14) × (28/25) = 140/350 = 2/5

Conditional Probability from Venn Diagrams (Higher)

Worked Example 7

In a Venn diagram, ξ has 50 elements. A has 30 elements, B has 25 elements, and A ∩ B has 10 elements.

Find: (a) $P(A\mid B)$ (b) $P(B\mid A)$

Solution:

(a) $P(A\mid B)$ = n(A ∩ B) / n(B) = 10/25 = 2/5

(b) $P(B\mid A)$ = n(A ∩ B) / n(A) = 10/30 = 1/3

Key Point: $P(A\mid B)$ and $P(B\mid A)$ are generally NOT equal. The order matters.

Worked Example 8 — Formula Applied Directly

$P(A)$ = 0.7, $P(B)$ = 0.4 and $P(A \cap B)$ = 0.28. Find:

(a) $P(A\mid B)$ (b) $P(B\mid A)$ (c) Are A and B independent? Justify.

Solution:

(a) Using $P(A\mid B) = \frac{P(A \cap B)}{P(B)}$ : $P(A\mid B)$ = 0.28 / 0.4 = 0.7

(b) Using $P(B\mid A) = \frac{P(A \cap B)}{P(A)}$ : $P(B\mid A)$ = 0.28 / 0.7 = 0.4

(c) $P(A\mid B)$ = 0.7 = $P(A)$ , so knowing B occurred does not change the probability of A. A and B are independent.

Worked Example 9 — Conditional Probability from a Venn Diagram

In a survey of 60 people, events M (likes maths) and S (likes science) are recorded:

n(M only) = 18
n(S only) = 14
n(M ∩ S) = 16
n(neither) = 12

Find: (a) $P(M\mid S)$ (b) $P(S\mid M)$ (c) $P(M'\mid S)$

Solution:

Total = 18 + 14 + 16 + 12 = 60 ✓

n(M) = 18 + 16 = 34, n(S) = 14 + 16 = 30, n(M ∩ S) = 16.

(a) $P(M\mid S)$ = n(M ∩ S) / n(S) = 16/30 = 8/15

Conditional Probability

Conditional Probability

What Is Conditional Probability?

Which Tool Should I Use?

Key Vocabulary

Conditional Probability from Two-Way Tables

Worked Example 1

Conditional Probability from Tree Diagrams

Worked Example 2

The Conditional Probability Formula

Worked Example 3

Testing for Independence Using Conditional Probability

Worked Example 4

Worked Example 5

Harder Conditional Probability Problems (Higher)

Worked Example 6

Conditional Probability from Venn Diagrams (Higher)

Worked Example 7

Worked Example 8 — Formula Applied Directly

Worked Example 9 — Conditional Probability from a Venn Diagram

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