Conditional Probability [H]

This lesson covers conditional probability — a Higher tier topic on the AQA GCSE Mathematics specification. Conditional probability deals with the probability of an event occurring given that another event has already occurred. You will learn to solve problems using tree diagrams, two-way tables, and Venn diagrams.

---

What Is Conditional Probability?

Conditional probability is the probability of an event happening, given that another event has already happened.

It is written as P(A | B), which reads as "the probability of A given B".

For example:

• P(rain | cloudy) = the probability of rain, given that it is cloudy.
• P(red second | red first) = the probability of picking a red counter second, given that a red counter was picked first.

Exam Tip: The key phrase to look for is "given that". Whenever you see it (or similar phrasing like "if", "knowing that", or "after"), you are dealing with conditional probability.

---

Conditional Probability with Two-Way Tables

A two-way table is one of the clearest methods for solving conditional probability questions.

Worked Example 1

100 students were surveyed about whether they study French or Spanish.

	French	Not French	Total
Spanish	25	30	55
Not Spanish	35	10	45
Total	60	40	100

(a) A student is chosen at random. Given that the student studies French, what is the probability that they also study Spanish?

We are told the student studies French, so we only consider the 60 students who study French.

Of these 60, how many also study Spanish? 25

P(Spanish | French) = 25/60 = 5/12

(b) Given that a student does not study Spanish, what is the probability that they study French?

We consider only the 45 students who do not study Spanish.

Of these 45, how many study French? 35

P(French | not Spanish) = 35/45 = 7/9

Exam Tip: When using a two-way table for conditional probability, the "given that" condition tells you which row or column to restrict your attention to. The total for that row or column becomes your new denominator.

---

Conditional Probability with Tree Diagrams

Tree diagrams for without replacement problems naturally involve conditional probability — the probabilities on the second set of branches depend on what happened first.

Worked Example 2

A box contains 6 chocolates and 4 toffees. Two sweets are chosen at random without replacement. Find the probability that the second sweet is a chocolate, given that the first sweet was a toffee.

After removing a toffee, the box contains: 6 chocolates and 3 toffees = 9 sweets total.

P(chocolate second | toffee first) = 6/9 = 2/3

Worked Example 3

Using the same box (6 chocolates, 4 toffees), find the probability that both sweets are the same type.

graph LR
    S[ ] -->|6/10 Choc| A[Choc]
    S -->|4/10 Toffee| B[Toffee]
    A -->|5/9 Choc| C[CC = 30/90]
    A -->|4/9 Toffee| D[CT = 24/90]
    B -->|6/9 Choc| E[TC = 24/90]
    B -->|3/9 Toffee| F[TT = 12/90]

P(both same) = P(CC) + P(TT) = 30/90 + 12/90 = 42/90 = 7/15

Exam Tip: In tree diagrams for without replacement, always update the numerator AND the denominator for the second set of branches. If a chocolate was taken first, there is one fewer chocolate AND one fewer sweet in total.

---

Using the Conditional Probability Formula [H]

The formal formula for conditional probability is:

P(A | B) = P(A and B) / P(B)

This formula is useful when you already know the individual probabilities.

Worked Example 4

In a school, P(student plays football) = 0.5, P(student plays cricket) = 0.3, and P(student plays both) = 0.15.

Find the probability that a student plays football, given that they play cricket.

P(football | cricket) = P(football and cricket) / P(cricket) = 0.15 / 0.3 = 0.5

Worked Example 5

P(A) = 0.6, P(B) = 0.5, P(A and B) = 0.2.

Find P(A | B) and P(B | A).

P(A | B) = P(A and B) / P(B) = 0.2 / 0.5 = 0.4

P(B | A) = P(A and B) / P(A) = 0.2 / 0.6 = 1/3

Note that P(A | B) is not the same as P(B | A).

---

Conditional Probability with Venn Diagrams [H]

Venn diagrams provide a visual method for conditional probability questions.

Worked Example 6

In a group of 50 people:

• 20 like tea
• 30 like coffee
• 10 like both

graph TD
    subgraph Universal Set - 50 people
        subgraph Tea - 20
            A[Tea only: 10]
            B[Both: 10]
        end
        subgraph Coffee - 30
            B
            C[Coffee only: 20]
        end
        D[Neither: 10]
    end

• Tea only = 20 - 10 = 10
• Coffee only = 30 - 10 = 20
• Neither = 50 - 10 - 10 - 20 = 10

(a) Given that a person likes tea, find the probability that they also like coffee.

We restrict attention to the 20 people who like tea. Of these, 10 also like coffee.

P(coffee | tea) = 10/20 = 1/2

(b) Given that a person likes coffee, find the probability that they also like tea.

We restrict to the 30 people who like coffee. Of these, 10 also like tea.

P(tea | coffee) = 10/30 = 1/3

Exam Tip: Note that P(coffee | tea) and P(tea | coffee) give different answers. Always make sure you identify the correct "given" group as your denominator.

---

Testing for Independence [H]

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

P(A | B) = P(A)

If knowing that B has happened does not change the probability of A, then A and B are independent.

Worked Example 7

Using the tea and coffee example above:

• P(tea) = 20/50 = 2/5
• P(tea | coffee) = 10/30 = 1/3

Since P(tea | coffee) is not equal to P(tea), the events "likes tea" and "likes coffee" are not independent.

---

Summary