Conditional probability | Pearson Edexcel GCSE Maths Notes

Visual model

Conditional probability changes the denominator

only use the restricted groupgiven B has happened

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

Gold-standard guide

26 mins

What you will learn

Calculate probability when outcomes depend on earlier information.

Use a clear step-by-step method for conditional probability.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier probability skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Conditional probability: P(B|A) = probability of B given A has occurred

Step 1

Find P(A and B)

P(A and B) = P(A) × P(B|A) = 0.5 × 0.6 = 0.3

Step 2

Find P(A' and B)

P(A' and B) = P(A') × P(B|A') = 0.5 × 0.2 = 0.1

Step 3

Find P(B) using the total probability rule

P(B) = P(A and B) + P(A' and B) = 0.3 + 0.1 = 0.4

Watch out

Students confuse P(A|B) with P(A and B)

Conditional

$P(A given B) = P(A and B) / P(B).$

Without replacement

Update the total after each $pic$ k.

Worked example

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

Find P(A and B): P(A and B) = P(A) × P(B|A) = 0.5 × 0.6 = 0.3.

Find P(A' and B): P(A' and B) = P(A') × P(B|A') = 0.5 × 0.2 = 0.1.

Find P(B) using the total probability rule: P(B) = P(A and B) + P(A' and B) = 0.3 + 0.1 = 0.4.

Final answer

P(A and B) = 0.3; P(B) = 0.4

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

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

3 marks4 minsconditional-probability-worked

Show solution

Worked solution

1.Find P(A and B): P(A and B) = P(A) × P(B|A) = 0.5 × 0.6 = 0.3.
2.Find P(A' and B): P(A' and B) = P(A') × P(B|A') = 0.5 × 0.2 = 0.1.
3.Find P(B) using the total probability rule: P(B) = P(A and B) + P(A' and B) = 0.3 + 0.1 = 0.4.

Final answer

P(A and B) = 0.3; P(B) = 0.4

Mark points

M1: find p(a and b)
M1: find p(a' and b)
M1: find p(b) using the total probability rule
A1: P(A and B) = 0.3; P(B) = 0.4

Watch out

Students confuse P(A|B) with P(A and B).P(B|A) is the probability of B assuming A has already happened — it is not the same as P(A and B).The formula is P(A and B) = P(A) × P(B|A), not P(B|A) alone.

Diagnosticrecall

P(A) = 0.4, P(B|A) = 0.7. Find P(A and B).

1 mark2 minsconditional-probability-q1

Show solution

Worked solution

1.Spot the skill: Conditional probability: P(B|A) = probability of B given A has occurred.
2.Use the find p(a and b) stage first, then find p(a' and b).
3.Keep the final answer visible: 0.28.

Final answer

0.28

Mark points

M1: use the correct conditional probability: p(b|a) = probability of b given a has occurred.p(a and b) = p(a) × p(b|a).use a tree diagram with p(a), p(a') on the first branches and p(b|a), p(b|a') on the second branches.
A1: 0.28

Watch out

Easyprocedure

Box 1 has 3 red, 2 blue. Box 2 has 1 red, 4 blue. A box is chosen at random, then a ball. Find P(red).

2 marks3 minsconditional-probability-q2

Show solution

Worked solution

1.Spot the skill: Conditional probability: P(B|A) = probability of B given A has occurred.
2.Use the find p(a' and b) stage first, then find p(b) using the total probability rule.
3.Keep the final answer visible: P(red) = 0.5×( $\frac{3}{5}$ ) + 0.5×( $\frac{1}{5}$ ) = 0.3 + 0.1 = 0.4.

Final answer

P(red) = 0.5×( $\frac{3}{5}$ ) + 0.5×( $\frac{1}{5}$ ) = 0.3 + 0.1 = 0.4

Mark points

M1: use the correct conditional probability: p(b|a) = probability of b given a has occurred.p(a and b) = p(a) × p(b|a).use a tree diagram with p(a), p(a') on the first branches and p(b|a), p(b|a') on the second branches.
A1: P(red) = 0.5×( $\frac{3}{5}$ ) + 0.5×( $\frac{1}{5}$ ) = 0.3 + 0.1 = 0.4

Watch out

Mediumreasoning

Using the data above: find P(box 1 | red).

3 marks4 minsconditional-probability-q3

Show solution

Worked solution

1.Spot the skill: Conditional probability: P(B|A) = probability of B given A has occurred.
2.Use the find p(b) using the total probability rule stage first, then find p(a and b).
3.Keep the final answer visible: P(box 1 and red)/P(red) = 0. $\frac{3}{0}$ .4 = $\frac{3}{4}$ .

Final answer

P(box 1 and red)/P(red) = 0. $\frac{3}{0}$ .4 = $\frac{3}{4}$

Mark points

M1: use the correct conditional probability: p(b|a) = probability of b given a has occurred.p(a and b) = p(a) × p(b|a).use a tree diagram with p(a), p(a') on the first branches and p(b|a), p(b|a') on the second branches.
A1: P(box 1 and red)/P(red) = 0. $\frac{3}{0}$ .4 = $\frac{3}{4}$

Watch out

Hardproblem solving

P(late | rain) = 0.8, P(rain) = 0.3. Find P(late and rain).

3 marks5 minsconditional-probability-q4

Show solution

Worked solution

1.Spot the skill: Conditional probability: P(B|A) = probability of B given A has occurred.
2.Use the find p(a and b) stage first, then find p(a' and b).
3.Keep the final answer visible: 0.24.

Final answer

0.24

Mark points

M1: use the correct conditional probability: p(b|a) = probability of b given a has occurred.p(a and b) = p(a) × p(b|a).use a tree diagram with p(a), p(a') on the first branches and p(b|a), p(b|a') on the second branches.
A1: 0.24

Watch out

Exam-stylemulti-step

Two events A and B are independent. P(A) = 0.4, P(B) = 0.5. Find P(A and B).

4 marks6 minsconditional-probability-q5

Show solution

Worked solution

1.Spot the skill: Conditional probability: P(B|A) = probability of B given A has occurred.
2.Use the find p(a' and b) stage first, then find p(b) using the total probability rule.
3.Keep the final answer visible: 0.2.

Final answer

0.2

Mark points

M1: use the correct conditional probability: p(b|a) = probability of b given a has occurred.p(a and b) = p(a) × p(b|a).use a tree diagram with p(a), p(a') on the first branches and p(b|a), p(b|a') on the second branches.
A1: 0.2

Watch out

Grade 9 stretchproblem solving

A bag has 5 red and 3 blue counters. Two are taken without replacement. Find P(at least one blue).

4 marks7 minsconditional-g9

Show solution

Worked solution

1.Use 1 - P(two red).
2.Multiply along the red-red route.
3.Subtract from 1.

Final answer

$\frac{9}{1}4$

Mark points

M1: 1 - ( $\frac{5}{8}$ × $\frac{4}{7}$ )
A1: $\frac{9}{1}4$

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

16 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Conditional probability - 2 marksP(A) = 0.4, P(B|A) = 0.7. Find P(A and B).Mark answer

Answer

0.28

2Probability basics - 2 marksP(A) = 0.35. Find P(not A).Mark answer

Answer

0.65

3Relative frequency - 2 marksExpected frequency of an outcome with probability 0.4 in 250 trials.Mark answer

Answer

100

4Sample spaces and frequency trees - 3 marksP(different colours) when drawing twice with replacement from the bag above.Mark answer

Answer

2 × ( $\frac{4}{6}$ × $\frac{2}{6}$ ) = $\frac{8}{1}8$ = $\frac{4}{9}$

Mastery check

I can explain the method for conditional probability.
I can show clear working without skipping key steps.
I can avoid this mistake: Students confuse P(A|B) with P(A and B).P(B|A) is the probability of B assuming A has already happened — it is not the same as P(A and B).The formula is P(A and B) = P(A) × P(B|A), not P(B|A) alone.

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