Conditional probability questions

A conditional probability is the probability of an event when you already know that another event has happened. "The probability of B given A" is written P(B | A). You find it by shrinking the set of outcomes to only those where A happened. The formula is P(B | A) = P(A and B) ÷ P(A). Below are worked conditional probability questions using two-way tables, Venn diagrams and tree diagrams, then questions to try.

The conditional probability formula

A ∩ B means "A and B". In words: of all the times A happens, what fraction also have B? When you are counting outcomes, the same idea is:

At GCSE (Higher tier) you calculate conditional probabilities from two-way tables, tree diagrams and Venn diagrams, often using expected frequencies. At A-level you also use the formula, which AQA prints in the specification.

Conditional probability from a two-way table

Example 1

200 people were asked whether they own a bike and whether they own a car.

• P(bike | car): only look at the 100 car owners. 34 of them own a bike, so P = 34/100 = 0.34.
• P(car | bike): only look at the 80 bike owners. 34 of them own a car, so P = 34/80 = 0.425.

The two answers are different. The "given" event decides which total you divide by.

Conditional probability from a Venn diagram

Example 2

• P(T | F): 18 + 7 = 25 students play football, and 7 of those also play tennis. P = 7/25 = 0.28.
• P(F | T): 7 + 13 = 20 play tennis, and 7 of those play football. P = 7/20 = 0.35.
• P(T | not F): 13 + 12 = 25 students do not play football, and 13 of them play tennis. P = 13/25 = 0.52.

There is more on reading Venn diagrams in Venn diagram rules.

Conditional probability on a tree diagram

On a tree diagram, the probabilities on the second set of branches are already conditional: each one is the probability of the second event given what happened on the first branch.

Example 3: without replacement

A bag holds 5 red counters and 3 blue counters. Two are taken out, one after the other, without replacement.

1. P(second is blue | first is red): after a red is taken, 7 counters are left and 3 are blue, so P = 3/7.
2. P(both the same colour) = P(red, red) + P(blue, blue) = (5/8 × 4/7) + (3/8 × 2/7) = 20/56 + 6/56 = 26/56 = 13/28.
3. P(both red | both the same colour) = P(red, red) ÷ P(same colour) = (20/56) ÷ (26/56) = 10/13.

The third part is the kind that catches people out: you are told the colours match, so the 26/56 becomes the whole of what is possible. Practise conditional probability in this question from the Probability course.

Using the formula (A-level)

Example 4: finding a conditional probability

P(A) = 0.6 and P(A ∩ B) = 0.15. Find P(B | A).

• P(B | A) = 0.15 ÷ 0.6 = 0.25.

Example 5: testing for independence

P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Are A and B independent?

• P(A | B) = 0.2 ÷ 0.5 = 0.4.
• This equals P(A), so knowing B happened does not change the probability of A: A and B are independent. (Equivalently, P(A) × P(B) = 0.2 = P(A ∩ B).)

Example 6: a screening test

2% of people have a condition. A test is positive for 95% of people who have it, and for 10% of people who do not. A person tests positive. What is the probability they have the condition?

1. P(condition and positive) = 0.02 × 0.95 = 0.019.
2. P(no condition and positive) = 0.98 × 0.10 = 0.098.
3. P(positive) = 0.019 + 0.098 = 0.117.
4. P(condition | positive) = 0.019 ÷ 0.117 = 0.162 (3 significant figures).

Only about 1 in 6 positives has the condition, because the condition is rare and the false positives come from the much larger group without it. The Probability Calculator helps you check combined probabilities like these.

Conditional probability questions to try

1. Using the bike and car table in Example 1, find P(no car | no bike).
2. Using the Venn diagram in Example 2, find P(F | not T).
3. A bag has 4 green and 6 yellow sweets. Two are eaten without replacement. Find P(second is green | first is yellow).
4. P(A) = 0.35, P(B) = 0.4 and P(A ∩ B) = 0.12. Find P(A | B), and say whether A and B are independent.
5. In Example 3, find P(first is blue | both the same colour).

Answers

1. 120 people have no bike and 54 of them have no car: 54/120 = 0.45.
2. 13 + 12 = 25 do not play tennis and 18 of those play football: 18/25 = 0.72.
3. After a yellow, 9 sweets remain and 4 are green: 4/9.
4. 0.12 ÷ 0.4 = 0.3. This is not equal to P(A) = 0.35, so A and B are not independent.
5. Both blue has probability 6/56, so P = (6/56) ÷ (26/56) = 3/13.

Common mistakes

• Dividing by the grand total. In P(B | A), divide by the number (or probability) in A, not by everyone.
• Mixing up P(B | A) and P(A | B). They are usually different; the event after the bar is the one you know happened.
• Using unchanged probabilities without replacement. After the first pick, both the total and the count change.
• Multiplying when you should divide. P(A and B) multiplies along branches; P(B | A) divides by P(A).
• Assuming independence. Check it: A and B are independent only if P(A | B) = P(A).

Practise with a short timed paper

Conditional probability is Higher tier content at GCSE (AQA P9) and is in A-level Maths statistics (AQA M2). The daily GCSE Maths Higher and A-level Maths: Statistics papers are short timed papers of 8 questions, with a new one every day.

Read next

Maths · ProbabilityHow to find probability: formula, rules and examplesHow to find probability step by step: the basic formula, not, or and and rules, tree diagrams, experimental probability and worked examples.